Em561 Lecture Notes - Part 3 Of 3[1] 3k93x

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8-1 Introduction • In the previous chapter we illustrated how a parameter can be estimated from sample data. • Estimate mean length of adult foot. Measure 28 students and find X = 26.322 cm. Is this a good estimate for μ? • Do we believe μ EXACTLY equals 26.322 cm (μ=X)?

• Rather than estimate μ with a single value (called point estimate) let‟s develop range or interval of values (called interval estimate) we think contains true value of μ. • Now rather than saying μ = 26.322 cm, we say 25.77 ≤ μ ≤ 26.87 cm

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8-1 Introduction • A confidence interval for an unknown parameter θ is an interval that contains a set of plausible or believable values of the parameter. • It is associated with a confidence level, 1-α, which measures the probability that the confidence interval actually contains the unknown parameter. • A 100(1-α) percent confidence interval on the unknown parameter θ: P(lower limit ≤ θ ≤ upper limit) = 1-α

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8-1 Introduction • For example, a 99% confidence interval on the parameter θ • Find the lower and upper limit such that: P(lower limit ≤ θ ≤ upper limit) = 0.99

• Common 100(1-α)% CI: • 99% CI → 1-α = 0.99 → α = 0.01 • 95% CI → 1-α = 0.95 → α = 0.05 • 90% CI → 1-α = 0.90 → α = 0.10

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8-1 Introduction • We use a sample statistic to estimate the population parameter • X to estimate μ, S2 to estimate σ2

• General procedure to form confidence interval – Take a sample – Find the sample statistic of interest (i.e. X or S2) – Form interval around sample statistic that we think contains true population parameter value • Calculate upper and lower CI limit

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8-1 Introduction • How to calculate upper and lower CI limits? • Using the sampling distribution of the appropriate sample statistic, choose CI such that: P(lower limit ≤ θ ≤ upper limit) = 1-α

• Use Normal distribution to find CI for μ when n≥40 • Use Student t distribution to find CI for μ when n<40 • Use Chi-square distribution to find CI for σ2

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8-1 Introduction • How to calculate upper and lower CI limits? P(lower limit ≤ θ ≤ upper limit) = 1-α

• As we will soon see, width (upper limit – lower limit) of a CI is a function of: • Confidence level, 1-α • As confidence level ↑, CI width ↑

• Sample size (i.e. # observations) • As sample size↑, CI width ↓

• Variance of data (as measured by s2) • As variance ↑, CI width ↑

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8-1 Introduction • The width of a confidence interval is a measure of the precision of estimation. • The width of the confidence interval is a measure of the quality of information obtained from the sample. • The wider the confidence interval, the more confident we are that the interval actually contains the true value of θ. – : as confidence level ↑, CI width ↑

• However, the wider the confidence interval the less information we have about the true value of θ • In an ideal situation, we obtain a relatively narrow interval with high confidence. This is possible when sample size is large and/or variance is small.

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8-1 Introduction • Precision of CI • Do not want CI that is too narrow • Low probability it will contain true population parameter

• Do not want CI that is too wide • Not useful to make decisions

• Mean length of foot example • 50% CI:

26.22 ≤ μ ≤ 26.42 cm

• 99% CI:

16.0 ≤ μ ≤ 36.0 cm

• 99% CI:

25.92 ≤ μ ≤ 26.72 cm

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8-1 Introduction • Two-sided confidence interval specifies both a lower and upper limit on θ P(lower limit ≤ θ ≤ upper limit) = 1-α • One-sided confidence interval may be more appropriate for some applications. • One-sided lower confidence interval on θ P(lower limit ≤ θ) = 1-α • One-sided upper confidence interval on θ P(θ ≤ upper limit) = 1-α

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8-1 Introduction • We will find confidence intervals for several population parameters: – Mean, μ, of any distribution, large sample size (n ≥ 40) – Mean, μ, of normal distribution, small sample size (n < 40) – Variance, σ2, of normal distribution – Population proportion, p

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8-2 Confidence Interval on the Mean, μ, for Large Sample

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• Want to estimate the population mean, μ, of data from any distributed with unknown variance, σ2 • Take a sample of size n ≥ 40 (Note: data can have any distribution) • Find sample mean, X , and sample variance, S2

• X will be Normally distributed with mean μ and variance σ2/n – How do I know this? Central Limit Theorem from Chpt 7. As sample size, n, gets large, the distribution of sample means approaches a normal distribution

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Distribution Plot

Density

 2  X ~ N   ,  n   X ~ ??(  , 2 )

μ • Data from UNKNOWN distribution with UNKNOWN mean, μ, and UNKNOWN variance σ2 • By CLT, sample means normally distributed with mean, μ, and variance σ2/n • Both distributions have same mean, μ, what is it? Form interval we think contains μ

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• Use sample mean, X , to estimate μ • Sample mean, X , will be center of CI for μ • Calculate upper and lower CI limit using X ~ N(μ,σ2/n) and 1-α such that P(lower limit ≤ μ ≤ upper limit) = 1-α • How to calculate lower limit and upper limit?

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f(x)

 σ2  X ~ N μ,   n 

X

Xμ

X

X

• Our sample mean is a single draw from this distribution. • We do not know which part of the curve our sample mean was drawn from. So we do not know how close the sample mean is to μ

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• All samples from this distribution will have same width CI. What should that width be?

 σ2  X ~ N μ,   n  f(x)

• We want the CI we form around our sample mean to have a high (1-α) probability of including μ

301

X

X

X

μ

X

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• We choose the CI limits such that P(lower limit ≤ μ ≤ upper limit) = 1-α • We choose upper and lower limits such that repeated sampling will result in 100(1-α)% of the CI containing μ – 95% CI example: choose upper and lower limit such that when we repeatedly take a sample of size n and form a CI, 95% of these CIs will contain μ

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How wide to make this interval so that it includes μ 100(1-α)% of the time?

Figure 8-1 Repeated construction of a confidence interval for .

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f(x)

 σ2  X ~ N μ,   n 





1-α

2

2

X μ

• 100(1-α)% of our samples will have a sample mean in this range. • If we choose the upper and lower CI limits so that these sample‟s CIs include μ, then 100(1-α)% of the CI will include μ


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 σ2  X ~ N μ,   n  f(x)

For example, 95% of samples will have sample mean in this range

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0.025

0.95

0.025 X

X Upper Limit

μ

X Lower Limit

Choose the CI width so that all sample means in this range will include μ. Consider most extreme cases

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• Using standard normal distribution, N(0,1), we find 95% of area under curve within ± 1.96 standard deviations of mean • So, for X ~N(μ,σ2/n), 95% of area under the curve within ± 1.96 s.d. of mean where s.d. = σ n  σ2  X ~ N μ,   n 

0.4

f(x)

f(x)

Z~N(0,1)

0.3

0.2

0.95

0.025

0.1

0.025

0.95

μ

Z

0.0

-1.96

0

0.025

X

0.025

1.96

1.96



1.96

n

• So our 95% CI on μ has a width of 2(1.96)(σ

 n

n)

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 σ2  X ~ N μ,   n 

f(x)

f(x)

 σ2  X ~ N μ,   n 

0.95

0.025



0.025

2

μ X  1.96

n

95% CI

X  1.96

 n

2

X μ

X





1-α

X X  Z / 2

X

 n

X  Z / 2

 n

100(1-α)% CI

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• 100(1-α)% Confidence Interval on μ (2-sided):

X  Z / 2

 n

   X  Z / 2

 n

• But we have unknown variance, σ2 • When large sample size (n ≥ 40), we can replace σ2 with sample variance, S2 • 100(1-α)% Confidence Interval on μ (2-sided):

X  Z / 2

S n

   X  Z / 2

S n

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8-2 Confidence Interval on the Mean, μ, for Large Sample • 100(1-α)% Confidence Interval on μ (2-sided):

X  Z / 2

S n

   X  Z / 2

S n

X  Z / 2

S n

• where Zα/2 is the upper 100(α/2) percentage point of the standard normal distribution • Zα/2 = z for which P(Z>z) = α/2 • For a 90% CI, α = 0.10 and Z0.05 = 1.645 • For a 95% CI, α = 0.05 and Z0.025 = 1.96 • For a 99% CI, α = 0.01 and Z0.005 = 2.576

These Z values come from the N(0,1) table

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8-2 Confidence Interval on the Mean, μ, for Large Sample • 100(1-α)% Confidence Interval on μ (2-sided):

X  Z / 2

S n

   X  Z / 2

S n

X  Z / 2

S n

• Notice: CI centered around sample mean • Notice: CI width is a function of confidence level, sample size, and variance

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• Kellogg‟s is interested in estimating the mean fill weight of their corn flakes cereal boxes. • They take a sample of 60 boxes and calculate the sample mean to be 12.05 oz. and the sample standard deviation to be 0.14 oz. 1. Find a 90% CI for the mean fill weight, μ •

Believable μ = 12.00 oz? μ = 12.10 oz? μ = 12.025 oz?

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One-sided Confidence Intervals • Often only interested in upper or lower CI limit (not both) – Lower limit of concrete strength – Upper limit on automated fill volume

• One-sided lower confidence interval on θ P(lower limit ≤ θ) = 1-α • One-sided upper confidence interval on θ P(θ ≤ upper limit) = 1-α


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One-sided lower confidence interval on μ • We choose the Lower CI limit such that P(lower limit ≤ μ) = 1-α • We choose the lower limit such that repeated sampling will result in 100(1-α)% of the one-sided lower CI containing μ • 95% CI example: choose lower limit such that when we repeatedly take a sample of size n and form a one-sided lower CI, 95% of these Lower CIs will contain μ • Only those 5% of sample means that are very large will not include μ in their Lower CI

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 σ2  X ~ N μ,   n 

95% of samples will have sample mean in this range

f(x)

Example of 95% Lower CI

0.95

0.05

X μ

X

Lower Limit

Choose the CI lower limit so that all sample means in this range will include μ. Consider most extreme case


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• Using standard normal distribution, N(0,1), we find 95% of area under curve is less than +1.645 standard deviations • So, for X ~N(μ,σ2/n), 95% of area under the curve is less than 1.645 s.d. of mean where s.d. =  n  σ2  X ~ N μ,   n 

0.4

Z~N(0,1)

Density

f(x)

0.3

0.2

0.95

0.1

0.95

X Z

0.0

0

0.05

0.05

μ

1.645

1.645

 n

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8-2 Confidence Interval on the Mean, μ, for Large Sample • 100(1-α)% Lower Confidence Bound on μ (1-sided):

X  Z 

S  n

• One-sided Upper CI calculated in much the same way as One-sided Lower CI • 100(1-α)% Upper Confidence Bound on μ (1-sided):

  X  Z 

S n

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• Kellogg‟s is interested in estimating the mean fill weight of their corn flakes cereal boxes. • They take a sample of 60 boxes and calculate the sample mean to be 12.05 oz. and the sample standard deviation to be 0.14 oz. 2. Find a 90% Lower Confidence Bound for the mean fill weight, μ •

Believable μ > 12.00 oz? μ > 12.10 oz?

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Summary of Procedure •

Data Distribution: Any i.e. X~ ??(μ, σ2)

•

Population parameters: – Mean, μ: unknown → to be estimated – Variance, σ2: unknown → estimate using S2

•

Sample Size: large i.e. n ≥ 40

•

Sample Statistic: Sample Mean

•

Sampling Distribution: Normal Distribution

•

CIs:

2-sided CI

X  Z / 2

S n

Upper 1-sided CI

  X  Z 

S n

Lower 1-sided CI

X  Z 

S  n

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8-2 Confidence Interval on the Mean, μ, for Large Sample How to do CI on μ, for large sample, in Minitab? • Can either enter each observation in a worksheet column or input summarized data Stat Basic Statistics 1-sample Z

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8-2 Confidence Interval on the Mean, μ, for Large Sample How to do CI on μ, for large sample, in Minitab?

n

Default is 95% two-sided CI

X

„Options‟ to change confidence level and one-sided CI

σ

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8-2 Confidence Interval on the Mean, μ, for Large Sample One-Sample Z The assumed standard deviation = 100

N

Mean

SE Mean

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215.0

24.3

95% CI (167.5, 262.5)

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8-2 Confidence Interval on the Mean, μ, for known population variance, σ2 • Use Normal distribution to find CI for μ when population variance, σ2, known

2-sided CI

X  Z / 2

Upper 1-sided CI

 n

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  X  Z / 2

 n

Lower 1-sided CI

X  Z / 2

 n



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8-3 Confidence Interval on the Mean, μ, of Normal Distribution with Small Sample

• Want to estimate the population mean, μ, of data that is Normally distributed with unknown variance (i.e. small sample) • Take a sample of size n from Normally distributed data (Note: n < 40) • Find sample mean, X •

X-

will be Student-t distributed with n-1 degrees of freedom

S n – How do I know this? from Chpt 7: Student t Distribution is sampling distribution of the sample mean when sample size, n, is small and underlying distribution is Normal (or close to Normal)

• Calculate upper and lower CI limit using Student-t distribution and 1-α such that P(lower limit ≤ μ ≤ upper limit) = 1-α

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8-3 Confidence Interval on the Mean, μ, of Normal Distribution with Small Sample • : Student-t distribution similar to Standard Normal Distribution, N(0,1) – Both symmetric, bell-shaped curves – Both have a mean of zero, μT =0, μZ = 0 – We use tables (t-tables or Z-table) to evaluate the area under the curve for standardized values • Example: standardized sample means

Z

X 



T

n

X  S

n


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• Using Student-t distribution, we find 100(1-α)% of area under curve within ± tα/2,n-1 of mean

f(x)

f(x)

• So for X, 100(1-α)% of area under the curve within ± tα/2,n-1 s.d. of mean where s.d. = S n

1-α

α/2

α/2

t -tα/2,n-1

0

tα/2,n-1

1-α

α/2

  t / 2,n 1 S

n

μ

α/2

  t / 2,n1 S

X n

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• 100(1-α)% Confidence Interval on μ (2-sided):

X  t / 2,n 1

S n

S

   X  t / 2,n 1

n

X  t / 2, n 1 

S n

• where tα/2,n-1 is the upper 100(α/2) percentage point of the student-t distribution with n-1 d.f. – Use t table in back of book to find appropriate t value for n and α

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8-3 Confidence Interval on the Mean, μ, of Normal Distribution with Small Sample • 100(1-α)% Upper Confidence Bound on μ (1-sided):

  X  t ,n 1

S n

• 100(1-α)% Lower Confidence Bound on μ (1-sided):

X  t ,n 1

S n



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8-3 Confidence Interval on the Mean, μ, of Normal Distribution with Small Sample • Suppose I measure my systolic blood pressure 5 times and get the following values: 98, 117, 132, 105, 121 • Assume my systolic blood pressure is normally distributed 1. Find a 95% CI on my mean systolic blood pressure • Believable my mean systolic BP is 100?

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8-3 Confidence Interval on the Mean, μ, of Normal Distribution with Small Sample 2. Believable my mean systolic BP is less than 100? Find appropriate 95% CI.

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Summary of Procedure •

Data Distribution: Normal i.e. X~ N(μ, σ2)

•

Population parameters: – Mean, μ: unknown → to be estimated – Variance, σ2: unknown → estimate using S2

•

Sample Size: small (i.e. n< 40)

•

Sample Statistic: Sample Mean

•

Sampling Distribution: Student-t Distribution

•

CIs:

2-sided CI

X  t / 2, n 1 

Upper 1-sided CI

S n

  X  t ,n 1

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S n

Lower 1-sided CI

X  t ,n 1

S n



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How to do CI on μ, with n small, in Minitab? • Can either enter each observation in a worksheet column or input summarized data  Stat Basic Statistics 1-sample t

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How to do CI on μ, with n small, in Minitab?

Default is 95% two-sided CI „Options‟ to change confidence level and one-sided CI

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One-Sample T: BP

Variable

N

Mean

StDev

SE Mean

BP

5

114.60

13.39

5.99

95% CI (97.97, 131.23)

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Confidence Intervals on μ • We discussed several different CI on μ – how to know which CI is appropriate for a particular problem? • Depends on data distribution, whether population variance, σ2, is known and sample size data distribution

σ2

sample size

CI distribution

any distribution

unknown

large

Z

Normal Distribution

unknown

small

t

Normal Distribution

known

any

Z

• 2-sided or 1-sided CI?

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Stop here on Saturday? • Turn in HW#1 • Review HW#1 solutions • Test #1 on Monday!

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8-4 Confidence Interval on the Variance, σ2, of a Normal Distribution • Want to estimate the population variance, σ2, of data that is Normally distributed • Take a sample of size n from Normally distributed data (Note: n is any size) • Find sample variance, S2 • S2 will be Chi-squared distributed (multiplied by a constant) with n-1 degrees of freedom – How do I know this? from Chpt 7: Chi-square Distribution is sampling distribution of the sample variance,S2

• Calculate upper and lower CI limit using S2 ~ (σ2/n-1)X2n-1 and 1-α such that P(lower limit ≤ σ2 ≤ upper limit) = 1-α

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8-4 Confidence Interval on the Variance, σ2, of a Normal Distribution

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• 100(1-α)% of our samples will have a Chi-squared value, X2, in this range where X2 = (n-1)S2/σ2

f(x)

 2  2   n 1 S 2 ~   n 1

α/2

1-α

α/2

0



2 1 / 2, n 1

  / 2,n1

(n  1) S 2

2

2

• Using Chi-squared distribution, we find 100(1-α)% of area under curve in range X21-(α/2),n-1 to X2α/2,n-1

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8-4 Confidence Interval on the Variance, σ2, of a Normal Distribution • 100(1-α)% Confidence Interval on σ2 (2-sided):

(n  1) S 2

  / 2,n1 2

  2

(n  1) S 2

12 / 2,n1

• Notice: CI width is a function of confidence level, sample size, and sample variance • Use Chi-squared table in back of book to find appropriate Chi-squared value for n and α • To find CI on σ, take square root of both CI limits

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8-4 Confidence Interval on the Variance, σ2, of a Normal Distribution • 100(1-α)% Upper Confidence Bound on σ2 (1-sided):

  2

(n  1) S 2

12 ,n1

• 100(1-α)% Lower Confidence Bound on σ2 (1-sided):

(n  1) S 2

  ,n 1 2

2

• To find CI on σ, take square root of CI limits

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8-4 Confidence Interval on the Variance, σ2, of a Normal Distribution • A rivet is to be inserted into a hole. A random sample of n = 15 parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is S = 0.008 mm. If the variance of the hole diameter is too large, an unacceptable proportion of rivets will not fit properly in the hole. Assume the hole diameter is normally distributed. • Construct a 99% upper confidence bound for σ2 • Believable σ < 0.010 mm? σ < 0.015 mm?

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Summary of Procedure • Data Distribution: Normal i.e. X~ N(μ, σ2) • Population parameters: – Variance, σ2: unknown → to be estimated

• Sample Size: any • Sample Statistic: Sample variance

• Sampling Distribution: Chi-squared Distribution • CIs: 2-sided CI

(n  1) S

2

  / 2,n1 2

2 

Upper 1-sided CI Lower 1-sided CI

(n  1) S



2

2 1 / 2, n 1

2 

(n  1) S 2



2 1 , n1

(n  1) S 2

  ,n 1 2

2

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How to do CI on σ2 in Minitab? • Can either enter each observation in a worksheet column or input summarized data Stat

Basic Statistics 1 Variance

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How to do CI on σ2, in Minitab? Choose to enter S or S2

n S

Default is 95% twosided CI „Options‟ to change confidence level and one-sided CI

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How to do CI on σ2, in Minitab?

Change to 99% CI Change to 1-sided upper bound CI

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8-4 Confidence Interval on the Variance, σ2, of a Normal Distribution Test and CI for One Standard Deviation Statistics N StDev 15 0.00800

Variance 0.000064

99% One-Sided Confidence Intervals

Method Standard

Upper Bound for StDev 0.01387

Upper Bound for Variance 0.000192

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8-5 Large Sample Confidence Interval for a Population Proportion • Suppose random sample of size n has been taken from a large population and X observations belong to a class of interest. – Example: take sample of 500 people. Count the number with long hair

• Sample proportion, pˆ = X/n, is point estimator of the proportion of the population that belongs to this class. • Can form CI around sample proportion

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8-5 Large Sample Confidence Interval for a Population Proportion • Want to estimate the population proportion, p, of data from any distribution • Take a large sample of size n from Normally distributed data (Note: n ≥40) • Find sample proportion, pˆ  • P will be normally distributed

X n

– How do I know this? See Chpt 4: Normal Approximation to the Binomial Distribution. As sample size gets large, binomial distribution can be approximated with Normal distribution.

• Calculate upper and lower CI limit using Normal Distribution and 1-α such that P(lower limit ≤ p ≤ upper limit) = 1-α

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8-5 Large Sample Confidence Interval for a Population Proportion • 100(1-α)% Confidence Interval on p (2-sided): pˆ  Z  / 2

pˆ (1  pˆ )  p  pˆ  Z  / 2 n

pˆ (1  pˆ ) n

pˆ  Z  / 2

pˆ (1  pˆ ) n

• where Zα/2 is the upper 100(α/2) percentage point of the standard normal distribution • Zα/2 = z for which P(Z>z) = α/2 • For a 90% CI, α = 0.10 and Z0.05 = 1.645 • For a 95% CI, α = 0.05 and Z0.025 = 1.96 • For a 99% CI, α = 0.01 and Z0.005 = 2.576

These Z values come from the N(0,1) table

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8-5 Large Sample Confidence Interval for a Population Proportion • 100(1-α)% Confidence Interval on p (2-sided): pˆ  Z  / 2

pˆ (1  pˆ )  p  pˆ  Z  / 2 n

pˆ (1  pˆ ) n

pˆ  Z  / 2

pˆ (1  pˆ ) n

• Notice: CI centered around sample proportion pˆ  • Notice: CI width is a function of confidence level, sample size, and variance – variance of binomial is

X n

npˆ (1  pˆ )

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8-5 Large Sample Confidence Interval for a Population Proportion • 100(1-α)% Upper Confidence Bound on p (1-sided):

p  pˆ  Z 

pˆ (1  pˆ ) n

• 100(1-α)% Lower Confidence Bound on p (1-sided):

pˆ  Z 

pˆ (1  pˆ ) p n

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8-5 Large Sample Confidence Interval for a Population Proportion A manufacturer of electronic calculators takes a random sample of 1200 calculators and finds there are 8 defective units. a) Construct a 95% confidence interval on the population proportion •

Believable that 1% defective?

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8-5 Large Sample Confidence Interval for a Population Proportion b) Is there evidence to a claim that the fraction of defective units produced is 1% or less?

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8-5 Large Sample Confidence Interval for a Population Proportion

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Summary of Procedure • Data Distribution: Any i.e. X~ ??(μ, σ2) • Population parameters: – Proportion, p: unknown → to be estimated

• Sample Size: large (n ≥ 40) • Sample Statistic: Sample proportion

• Sampling Distribution: Normal Distribution • CIs: 2-sided CI

pˆ  Z  / 2

pˆ (1  pˆ ) n

Upper 1-sided CI p  pˆ  Z 

pˆ (1  pˆ ) n

Lower 1-sided CI pˆ  Z 

pˆ (1  pˆ ) p n

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8-5 Large Sample Confidence Interval for a Population Proportion How to do CI on p, in Minitab? • Can either enter each observation in a worksheet column or input summarized data Stat

Basic Statistics 1 Proportion

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8-5 Large Sample Confidence Interval for a Population Proportion How to do CI on p, in Minitab?

X n

Default is 95% two-sided CI „Options‟ to change confidence level and one-sided CI

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8-5 Large Sample Confidence Interval for a Population Proportion Test and CI for One Proportion

Sample

X

N

Sample p

95% CI

1

8

1200

0.006667

(0.002882, 0.013094)

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Confidence Intervals • We discussed several different CI – how to know which CI is appropriate for a particular problem? • What population parameter is being estimated? μ? σ2? p? – If μ being estimated, is σ2 known? Sample size? • If σ2 known, then use Z • If σ2 unknown and large sample, then use Z

• If σ2 unknown and small sample, then use t – If σ2 being estimated, use Chi-squared – If p being estimated, use Z

• 2-sided or 1-sided CI?

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More Chapter 8 Examples • A drink machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 drinks were found to have a mean syrup content of 1.10 fluid ounces and a standard deviation of 0.015 fluid ounces. • Believable that the mean syrup dispensed is 1.15 fl oz? 1.05 fl oz? Find a 95% CI.

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More Chapter 8 Examples • The percentage of titanium in an alloy used in aerospace castings is measured in 51 randomly selected parts. The sample standard deviation is S = 0.37 percent. • Believable that the standard deviation is 0.50 percent? Find a 95% CI. • Believable that the variance is 0.13 percent? Find a 95% CI.

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More Chapter 8 Examples • Need more Chapter 8 examples? • READ THE BOOK! • We covered all sections 8-1 thru 8-6 (not 87)

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9-1 Hypothesis Testing • In previous chapter we learned to construct CI estimate of population parameter from sample data • In this chapter we‟ll learn how to accept or reject a statement or hypothesis about a parameter – Useful since we can formulate many types of decisionmaking problems, tests, or experiments in the engineering world as hypothesis-testing problems

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CI and Hypothesis Testing • Both CI (Chpt 8) and Hypothesis testing (Chpt 9) are used to make a statistical inference – Use sample statistic to „guess‟ population parameter

• In CI we formed an interval around sample statistic and used the interval to answer questions – Believable that μ=12? Believable that σ<3.2?

• In hypothesis testing we are making a statement about the population parameter then using the sample statistic to decide whether the statement is believable or not

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9-1 Hypothesis Testing • Suppose we are interested in the average length of an adult‟s foot • Take sample • Can use sample to form CI we think contains population mean, μ • OR can use sample to decide whether a statement or hypothesis about the population mean is believable • e.g. Hypothesis is μ = 27.0 cm

• In both CI and hypothesis testing we use the sample statistic (e.g. sample mean) to make inference about the unknown population statistic (e.g. μ)

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9-1 Hypothesis Testing • State our hypothesis in of Null Hypothesis, H0 and Alternative Hypothesis, H1 Two-sided hypothesis H0: μ = 27.0 cm H1: μ ≠ 27.0 cm One-sided hypotheses H0: μ ≥ 27.0 cm H1: μ < 27.0 cm

H0: μ ≤ 27.0 cm OR

H1: μ > 27.0 cm

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9-1 Hypothesis Testing • General hypotheses for population mean μ Two-sided hypothesis H0: μ = μ0 H1: μ ≠ μ0 One-sided hypotheses H0: μ ≥ μ0 H1: μ < μ0

H0: μ ≤ μ0 OR

H1: μ > μ0”

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9-1 Hypothesis Testing • Use sample statistics to decide which statement, H0 or H1, is more believable

Example

H0: μ = 27.0 cm H1: μ ≠ 27.0 cm

• If sample mean = 26.8 cm I would believe H0: μ = 27.0 cm • If sample mean = 23.1 cm I would believe H1: μ ≠ 27.0 cm

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9-1 Hypothesis Testing • How close does sample mean need to be to 27.0 cm for me to believe H0: μ = 27.0 cm? • How far away from 27 cm does the sample mean need to be for me to believe H1: μ ≠ 27.0 cm? • Depends on: • Variance, σ2 • Sample size, n • Significance level, α

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9-1 Hypothesis Testing • Example: H0: μ = 27.0 Reject H0 μ ≠ 27.0

H1: μ ≠ 27.0 Reject H0 μ ≠ 27.0

Accept H0 μ = 27.0

26.0

27.0

28.0

X

• If 26.0 ≤ sample mean ≤ 28.0, Accept H0 • If sample mean < 26.0 or if sample mean > 27.0, Reject H0

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9-1 Hypothesis Testing • Example: H0: μ = 27.0 Rejection Region Reject H0 μ ≠ 27.0

H1: μ ≠ 27.0

Acceptance Region Accept H0 μ = 27.0

26.0

27.0 Critical values

Rejection Region Reject H0 μ ≠ 27.0

28.0

X

How to choose critical values?

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9-1 Hypothesis Testing • How to choose critical values? • Use the sampling distribution of the appropriate sample statistic to determine critical values and therefore whether to accept or reject H0 • Use Normal distribution to find CI for μ when n≥40 • Use Student t distribution to find CI for μ when n<40 • Use Chi-square distribution to find CI for σ2

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9-1 Hypothesis Testing • We either Accept or Reject H0 • Always in of accepting/rejecting H0 , NOT H1

• When we reject H0 we are saying H0 is not plausible or believable • When we accept H0 we are saying H0 is plausible or believable. However accepting H0 does not prove H0 to be true. – There may be other plausible H0

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9-1 Hypothesis Testing • The “strongest” inference is available when the null hypothesis is rejected. • This point is important when an experimenter decides what should be the null hypothesis and what should be the alternative hypothesis for onesided problems.

• In order to “prove” or establish a statement (say μ > μ0) it is necessary to make it the alternative hypothesis (H1). • For one-sided hypothesis test, make H1 whatever you are trying to „prove‟

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9-1 Hypothesis Testing • We either Accept or Reject H0 based on our sample statistic • Our decision to accept or reject H0 will either be right or wrong. • We‟ll never know for certain whether we‟re right or wrong (because we do not know population parameter), but we can find the probability we‟re wrong • Two ways we could be wrong: – Reject H0 when H0 is true (type I error) – Accept H0 when H0 is false (type II error)

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9-1 Hypothesis Testing

• Probability of Type I error = α • α also called significance level

• Probability of Type II error =β • 1-β called power of the hypothesis test

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9-1 Hypothesis Testing

• Power, 1-β, can be interpreted as the probability of correctly rejecting a false null hypothesis.

• We often compare statistical tests by comparing their power properties.

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9-1 Hypothesis Testing General procedure for hypothesis test 1. From the problem context, identify the parameter of interest. 2. State the null hypothesis, H0 . 3. Specify an appropriate alternative hypothesis, H1. 4. Choose a significance level, . 5. Determine an appropriate test statistic. 6. State the rejection region for the statistic. 7. Calculate test statistic. 8. Decide to accept or reject H0 and report conclusion in the context of the problem

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9-1 Hypothesis Testing • We will conduct hypothesis tests for several population parameters: – Mean, μ, of any distribution, large sample size (n ≥ 40) – Mean, μ, of normal distribution, small sample size (n < 40) – Variance, σ2, of normal distribution – Population proportion, p

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9-2 Hypothesis Test on the Mean, μ, for Large Sample Null Hypothesis: H0: μ = μ0 Test Statistic: Z 0 

X  0 S n

380

• • • •

State hypothesis Take sample Calculate test stat Compare to appropriate rejection criteria • Accept or reject H0

Alternative Hypothesis

Rejection criteria

H1 :   0 H1 :   0 H1 :   0

Z 0  z / 2 Z 0  z Z 0   z

Rejection criteria in box on p.307 are incorrect

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9-2 Hypothesis Test on the Mean, μ, for Large Sample

381

• Calculate Test Statistic (i.e. standardize sample mean), Z0

a) Reject H0: μ=μ0 (i.e. believe H1: μ≠μ0) if test statistic very large or very small b) Reject H0: μ=μ0 (i.e. believe H1: μ>μ0) if test statistic very large c) Reject H0: μ=μ0 (i.e. believe H1: μ<μ0) if test statistic very small

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How does this Hypothesis Test Work? • Suppose

H0: μ = 10.0

H1:μ ≠ 10.0

• We want to find critical values for sample of size 25 with σ=2 Rejection Region Reject H0 μ ≠ 10

Acceptance Region

Accept H0 μ = 10

???

10

Rejection Region Reject H0 μ ≠ 10

???

X

Critical values

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How does this Hypothesis Test Work? • For α=0.05, critical values are ± 1.96 standard deviations from μ0=10.0  • from CLT  X  n • So critical values are

  9.216 10  1.96 2  25  

  10.784 10  1.96 2  25  

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How does this Hypothesis Test Work? • Take a sample. If sample mean between 9.216 and 10.784 then we believe μ=10 at α=0.05


Acceptance Region Accept H0 μ = 10

9.216

10


10.784

X

Critical values

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How does this Hypothesis Test Work? • Similarly we can work with the standardized values.

X  0 Z0  / n Known population st. dev.

H0:μ=10 H1:μ≠10

Equivalent to

Z0 

X  0 S n

Large sample size (n≥40)

H0:Z0=0 H1:Z0≠0

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How does this Hypothesis Test Work? • Take a sample. Find test statistic, Z0 (i.e. standardize sample mean). If Z0 between -1.96 and 1.96 then we believe Z0=0 or equivalently μ=10 at α=0.05 Rejection Region Reject H0

Acceptance Region Accept H0

-1.96

0

Rejection Region Reject H0

1.96

Z0

Critical values

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• The percent yield of a chemical process is being studied. • 75 observations on yield have been taken and their sample mean is 90.68% and sample variance 11.24% • Is it believable the true mean yield is 90.0%? Use α=0.05

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Stop here Tuesday? • Chapter 8 extra credit

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How to do hypothesis test on μ, with σ2 known, in Minitab? • Can either enter each observation in a worksheet column or input summarized data Stat Basic Statistics 1-sample Z

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How to do hypothesis test on μ, with σ2 known, in Minitab?

n

Default is H1≠μ0

X

„Options‟ to change to H1>μ0 or H1<μ0

σ

μ0

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One-Sample Z Test of mu = 90 vs not = 90 The assumed standard deviation = 3

N

Mean

SE Mean

5

90.68

1.34

95% CI (88.05, 93.31)

Z

P

0.51

0.612

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P-value • The hypothesis test can quickly show decisions (accept or reject) for a variety of significance levels, α • We can calculate the p-value associated with a hypothesis test • P-value is the smallest level of significance (i.e. α) that would lead to rejection of the null hypothesis H0 with the given data.

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P-value • The P-value is the probability that the test statistic will take on a value that is at least as extreme as the observed value of the statistic when the null hypothesis H0 is true. • Thus, a P-value conveys much information about the weight of evidence against H0, and so a decision maker can draw a conclusion at any specified level of significance. • P-value indicates how strongly we believe H0 or H1

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P-value • We reject H0 when p-value < α • Suppose p-value = 0.0316 • At α=0.05 accept or reject H0? • At α=0.01 accept or reject H0?

• So when p-value very small (i.e. near 0) we STRONGLY reject H0. – We REALLY do not believe H0 – We REALLY do believe H1

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Relationship between Hypothesis Test and CI

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Close relationship between CI and hypothesis test • H0: μ = μ0 • H1: μ ≠ μ0 • For same sample and significance level,α, the H0 will be rejected if and only if μ is NOT in the 100(1α)% CI on μ • This relationship between hypothesis test and CI is same whether population parameter is μ,σ2, or p

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• For a specific significance level, α, a CI is more informative than performing a hypothesis test. • The decision made by the hypothesis test can be deduced from the confidence interval by looking to see whether μ0 is inside the confidence interval or not.

• Thus the confidence interval portrays the decisions made by hypothesis test for all possible values of μ0 (for a given α)

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• However, for a specific sample, a hypothesis test is more informative than performing a CI because the hypothesis test can quickly show decisions (accept or reject) for a variety of significance levels, α.

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9-3 Hypothesis Test on the Mean, μ, of Normal Distribution with Small Sample Null Hypothesis: H0: μ = μ0 Test Statistic: Alternative Hypothesis

H1 :   0 H1 :   0 H1 :   0

T0 

X  0 S n

• • • •


Rejection criteria

T0  t / 2,n 1 T0  t ,n 1 T0  t ,n 1

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9-3 Hypothesis Test on the Mean, μ, of Normal Distribution with Small Sample

Figure 9-8 The reference distribution for H0:  = 0 with critical region for (a) H1:   0 , (b) H1:  > 0, and (c) H1:  < 0. a) Reject H0: μ=μ0 (i.e. believe H1: μ≠μ0) if test statistic very large or very small b) Reject H0: μ=μ0 (i.e. believe H1: μ>μ0) if test statistic very large c) Reject H0: μ=μ0 (i.e. believe H1: μ<μ0) if test statistic very small

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9-3 Hypothesis Test on the Mean, μ, of Normal Distribution with Small Sample

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• The tar content in cigars is being studied. A sample of 30 cigars is taken and the sample mean is 1.529mg and S=0.0566mg. Assume the tar content is normally distributed. • Can you a claim the mean tar content exceeds 1.5mg? Use α=0.05

How can we test to see if data is normally distributed?

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9-4 Hypothesis Test on the Variance, σ2, of a Normal Distribution Null Hypothesis: H0:σ2=σ02 Test Statistic:   2 0

(n  1) S 2

 02

• • • •



Rejection criteria

H 1 :  2   02 H 1 :  2   02 H 1 :  2   02

 02   2 / 2,n 1 or  02  12 / 2,n 1  02   2 ,n 1 Rejection criteria in box  02  12 ,n 1 on p.323 are incorrect

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9-4 Hypothesis Test on the Variance, σ2, of a Normal Distribution

a) Reject H0: σ2=σ02 (i.e. believe H1: σ2≠σ02 ) if test statistic very large or very small b) Reject H0: σ2=σ02 (i.e. believe H1: σ2>σ02 ) if test statistic very large c) Reject H0: σ2=σ02 (i.e. believe H1: σ2<σ02 ) if test statistic very small

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9-4 Hypothesis Test on the Variance, σ2, of a Normal Distribution • An engineer is testing the tire life for a new rubber compound. 16 tires were tested to end-of-life on a road test. The sample mean is 60139.7 km and S = 3645.94 km. Assume the tire life is normally distributed. • Can you conclude that the standard deviation of tire life exceeds 3500 km? Use α=0.05

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9-5 Hypothesis Test on a Population Proportion Null Hypothesis: H0: p = p0 Test Statistic: Z 0 

X  np0 np0 (1  p0 )

• • • •



Rejection criteria

H1: p ≠ p0

|Z0| > Z α/2

H1: p > p0

Z0 > Z α

H1: p < p0

Z0 < −Z α

Rejection criteria in box on p.326 are incorrect

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9-5 Hypothesis Test on a Population Proportion • The fraction of defective integrated circuits is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. • Do the data the claim the fraction of defective units produced is less than 0.05? Use α=0.05

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Hypothesis Testing • We discussed several different hypothesis tests – how to know which test is appropriate for a particular problem? • What population parameter is being estimated? μ? σ2? p? – If μ being estimated, is σ2 known? • If σ2 known or large sample, then use Z test statistic • If σ2 unknown and small sample, then use t test statistic – If σ2 being estimated, use Chi-squared test statistic – If p being estimated, use Z test statistic

• 2-sided or 1-sided hypothesis test?

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9-7 Testing for Goodness of Fit Is a particular distribution a good fit to the data? • H0: data is well fit by proposed distribution • H1: data is not well fit by proposed distribution 1. Divide proposed distribution into intervals a. # intervals ≥ 5 b. # expected observations in each interval ≥ 5

c. Therefore # observations ≥ 25 2. Count number of actual observations in each interval 3. Find number of expected observations in each interval for proposed distribution 4. Do hypothesis test to determine goodness of fit

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9-7 Testing for Goodness of Fit • The test is based on the Chi-square distribution. • Assume there is a sample of size n from a population whose probability distribution is unknown. • Let Oi be the observed frequency in the ith class interval. • Let Ei be the expected frequency in the ith class interval. • Test statistic:

(Oi  Ei ) 2   Ei i 1 k

2 0

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9-7 Testing for Goodness of Fit (Oi  Ei ) 2 • The test statistic is    Ei i 1 k

2 0

• Reject H0 if • Where

 02   2 , k  p 1

k = # intervals p = # distribution parameters estimated by sample statistics

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9-7 Testing for Goodness of Fit • Is it believable the following data (n=50) came from a U(36,78) distribution? 77.72 58.19 43.12 44.22 43.96 57.95 59.95 36.93 48.90 57.02

45.64 57.30 61.87 77.51 36.19 55.96 76.79 62.64 61.37 42.78

45.90 59.59 74.61 66.22 76.19 43.93 54.92 63.31 67.64 42.89

72.69 76.35 57.66 74.85 46.39 63.46 63.73 46.34 57.10 56.96

43.69 69.98 45.00 68.82 55.71 54.54 55.46 67.13 76.26 62.77

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9-7 Testing for Goodness of Fit H0: data well fit by U(36,78) H1: data not well fit by U(36,78) 1. Divide U(36,78) into intervals 2. Count # of actual observations in each interval 3. Find expected # observations in each interval for U(36,78) distribution 4. Do Goodness-of-Fit Hypothesis test

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9-7 Testing for Goodness of Fit 1. Divide U(36,78) into intervals 2. Count # of actual observations in each interval

interval

actual # obs

[36.0, 44.4)

9

[44.4, 52.8)

6

[52.8, 61.2)

14

[61.2, 69.6)

11

[69.6, 78.0]

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9-7 Testing for Goodness of Fit 3. Find expected # observations in each interval for U(36,78) distribution 1/42

36.0

44.4

78.0

•

P(36.0<X<44.4) = (44.4 – 36.0)(1/42) = 0.20

•

So, E = nP = 50(0.20) = 10

•

Repeat for each interval.

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9-7 Testing for Goodness of Fit

 20   i

 20 

interval

actual # obs (O)

Expected # obs (E)

[36.0, 44.4)

9

10

[44.4, 52.8)

6

10

[52.8, 61.2)

14

10

[61.2, 69.6)

11

10

[69.6, 78.0]

10

10

(Oi  E i ) Ei

(10  9) 2 (10  6) 2 (10  14) 2 (10  11) 2 (10  10) 2      3. 4 10 10 10 10 10

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9-7 Testing for Goodness of Fit • X02 = 3.4 • Critical value (at α=0.05) = X20.05,k-p-1 = X20.05,5-0-1=9.49 • Note: 0 distribution parameters estimated from data – we were told which distribution to try, U(36,78) • Since test statistic = 3.4 < critical value = 9.49, we ACCEPT H0 • We do believe a U(36,78) distribution is a good fit to this data

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9-7 Testing for Goodness of Fit • Number of customers arriving at an ATM is hypothesized to follow a Poisson distribution. • Data is collected over several days and the number of customers to arrive at the ATM each minute is summarized as follows # arrivals per minute 0 1 2 3 4 5

Observed Frequency 146 132 72 36 11 3

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9-7 Testing for Goodness of Fit • Estimate mean of hypothesized Poisson Distribution from the observed data • E(X) = λ = 1.1075 # arrivals per minute, x 0 1 2 3 4 5

: E(X)=∑xP(x) Observed Frequency 146 132 72 36 11 3 400

Observed P(X) 0.3650 0.3300 0.1800 0.0900 0.0275 0.0075 E(X)=1.1075

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9-7 Testing for Goodness of Fit • H0: Data is well fit by Poisson with λ=1.1075 • H1: Data is not well fit by Poisson with λ=1.1075 • Now calculate expected frequency (for Poisson with λ=1.1075) in each interval # arrivals per minute, x 0 1 2 3 4 ≥5

P(X) for Poisson with λ=1.1075 0.3304 0.3659 0.2026 0.0748 0.0207 0.0056

Expected Frequency with n=400 132.15 146.36 81.05 29.92 8.28 2.24

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9-7 Testing for Goodness of Fit • Since last interval has < 5 expected observations, combine it with previous interval # arrivals per minute, x 0 1 2 3 ≥4

P(X) for Poisson with λ=1.1075 0.3304 0.3659 0.2026 0.0748 0.0263

Expected Frequency with n=400 132.15 146.36 81.05 29.92 10.52

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9-7 Testing for Goodness of Fit • Now perform hypothesis test • • • • •

H0: Data is well fit by Poisson with λ=1.1075 H1: Data is not well fit by Poisson with λ=1.1075 Calculate test stat, X02 Compare to appropriate rejection criteria Accept or reject H0

# arrivals per minute 0 1 2 3 ≥4

Observed Frequency, O 146 132 72 36 14

Expected Frequency, E 133.82 146.53 80.22 29.28 10.15

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9-7 Testing for Goodness of Fit • X02 = 6.3950 • Critical value (at α=0.05) = X20.05,k-p-1 = X20.05,5-1-1=7.81 # arrivals per minute 0 1 2 3 ≥4

Observed Frequency, O 146 132 72 36 14

Expected Frequency, E 133.82 146.53 80.22 29.28 10.15

(Oi  Ei ) 2   Ei i 1 k

X02 1.1094 1.4405 0.8431 1.5413 1.4606 6.3950

2 0

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9-7 Testing for Goodness of Fit • Since test statistic X02 = 6.3950 < critical value X20.05,5-1-1=7.81, we ACCEPT H0

• We do believe a Poisson λ=1.1075 distribution is a good fit to this data

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More Chapter 9 examples • A particular type of gasoline is supposed to have a mean octane rating greater than 90%. Five measurements are taken of the octane rating as follows: 90.8%, 88.4%, 89.2%, 91.6%, 92.1% • Can you conclude that the mean octane rating is greater than 90%? Use α=0.10

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More Chapter 9 examples • A rivet is to be inserted into a hole. A random sample of n = 15 parts is selected, and the hole diameter is measured. • The sample standard deviation of the hole diameter measurements is S = 0.008 mm. If the variance of the hole diameter is too large, an unacceptable proportion of rivets will not fit properly in the hole. Assume the hole diameter is normally distributed. • Believable the population standard deviation less than 0.012 mm? use α=0.01

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More Chapter 9 examples • A new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specs call for the blocks to have a mean compressive strength greater than 1350 kPa. A sample of 100 blocks is produced and tested. The mean strength of the sample is 1360 kPa and the standard deviation is 70 kPa. • Is it believable the blocks meet spec? Use α=0.05

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Stop here Wednesday • Chapter 9 extra credit

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Regression Models •An experimenter is often interested in how a particular variable depends upon one or more of the other variables. •Modeling is often performed by finding a functional relationship between the expected value of a dependent variable, Y, and a set of independent variables, Xi. How does X affect Y? What is the effect of X on Y?

•Linear Regression is a modeling technique in which the expected value of a dependent variable is modeled as a linear combination of a set of independent variables. •Many problems in engineering and science involve exploring the relationships between two or more variables.

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Regression Analysis Example •A company that makes paper grocery bags is interested in improving the tensile strength of their bags. Specifically they are interested in the relationship between the hardwood concentration in the pulp and the tensile strength of the bag.

•Regression analysis can be used to build a model to relate tensile strength (Y) to hardwood concentration (X).

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Linear Regression Fit a simple linear regression model to paper bag example

Scatterplot of strength vs hardwood 25

strength

20

15

10

5 5.0

7.5

10.0

12.5 hardwood

15.0

17.5

20.0

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11-1 Empirical Models Based on the scatter diagram, it is probably reasonable to assume that the variable Y is related to X by the following simple linear regression model :

Y = β0 + β1X + ε where the slope, β1, and intercept, β0, of the line are called regression coefficients and where  is the random error term. We assume the mean and variance of  are 0 and 2

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11-2 Simple Linear Regression • The case of simple linear regression considers

a single regressor or predictor x and a dependent or response variable Y. • How to fit line to data? How to find values for slope, β1, and intercept, β0? •find slope, β1, and intercept, β0, of best fitting line.

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Simple Linear Regression • Used when demand is linearly increasing or decreasing over time

Demand

• Find „best‟ fitting line to data 95 85 75 65 55 45 35 25 0

5

10

15

20

25

Time

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11-2 Simple Linear Regression • The method of least squares is used to

estimate the parameters, 0 and 1 by minimizing the sum of the squares of the vertical deviations in Figure 11-3.

Figure 11-3 Deviations of the data from the estimated regression model.

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11-2 Simple Linear Regression • The sum of the squares of the deviations of the

observations from the true regression line is

• How to find 0 and 1 that minimize L?

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Simple Linear Regression

Fitted or estimated regression line is

yˆ  ˆ0  ˆ1 x

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Simple Linear Regression Fit a simple linear regression model to paper bag example Hardwood

Strength

x 5 5 5 5 5 5 10 10 10 10 10 10 15 15 15 15 15 15 20 20 20 20 20 20

y 7 8 15 11 9 10 12 17 13 18 19 15 14 18 19 17 16 18 19 25 22 23 18 20

x2 25 25 25 25 25 25 100 100 100 100 100 100 225 225 225 225 225 225 400 400 400 400 400 400

y2 49 64 225 121 81 100 144 289 169 324 361 225 196 324 361 289 256 324 361 625 484 529 324 400

xy 35 40 75 55 45 50 120 170 130 180 190 150 210 270 285 255 240 270 380 500 440 460 360 400

300 12.5

383 15.96

4500

6625

5310

Total Avg

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Simple Linear Regression ˆ1 

(383)(300) 24 300 2 4500  24

5310 

 0.697

ˆ0  15.96  0.697(12.5)  7.25

• Y = 7.25 + 0.697*X • Strength = 7.25 + 0.697*Hardwood

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Regression Analysis in Minitab

440

• Data in 2 columns: dependent (Y) & independent(X)  Stat  Regression Regression

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Regression Analysis in Minitab Regression Analysis: strength versus hardwood The regression equation is strength = 7.25 + 0.697 hardwood Predictor Constant hardwood S = 2.60201

Coef 7.250 0.69667

SE Coef 1.301 0.09501

R-Sq = 71.0%

Analysis of Variance Source DF SS Regression 1 364.01 Residual Error 22 148.95 Total 23 512.96

T 5.57 7.33

P 0.000 0.000

R-Sq(adj) = 69.6% MS 364.01 6.77

F 53.76

P 0.000

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Finding the regression equation in Excel

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• Put data in 2 columns – dependent (Y) – Independent (X)

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Regression in Excel • In Excel > Data Analysis >Regression

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Regression in Excel

strength = 7.25 + 0.697 hardwood

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Prediction of New Observations If x0 is the value of the regressor variable of interest,

Yˆ0  ˆ 0  ˆ 1 x0 is the point estimator of the new or future value of the response, Y0. Predict bag strength for a hardwood concentration of 8% Strength = 7.25 + 0.697(8) = 12.826 How did I know to use 8 and not 0.08?

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Prediction of New Observations A 100(1-α)% prediction interval on a future observation Y0 at the value x0 is given by Yˆ0  t  2

where

,N  2

  ( x0  x ) 2 2N 1 ˆ  2  N  x  2  x  N 

     

Yˆ0  ˆ 0  ˆ 1 x0

and where ˆ

2

Y 

2

 ˆ o  Y  ˆ 1  XY N 2

 MS E

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Prediction of New Observations Find a 95% PI for the tensile strength at a hardwood concentration of x0 = 8% Yˆ0  12.826

ˆ 2 

6625  7.25(383)  0.697(5310)  6.69 24  2

  25 (8  12.5) 2 12.826  2.074 6.69  2  24 4500  300  24

    

(7.28, 18.37)

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Regression Analysis in Minitab • Prediction intervals in Minitab – „Options‟ menu within Regression – Enter X0 value for PI

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Regression Analysis in Minitab Predicted Values for New Observations New Obs Fit SE Fit 95% CI 1 12.823 0.682 (11.409, 14.237)

95% PI (7.245, 18.402)

Values of Predictors for New Observations New Obs hardwood 1 8.00

Difference between CI and PI: CI refers to the true mean response at X0. CI based only on data used to fit the regression model. PI for future observation which is independent of observations used to develop regression model. Therefore more error (fitted model error and error associated with projecting into future) and wider interval.

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Linear Regression A few questions to answer about this linear regression analysis • How good a fit is this line to the data?

• Does the independent variable have a significant effect on the dependent variable? – Does X have a significant effect Y? – Does hardwood concentration have a significant effect on paper tensile strength?

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Adequacy of the Regression Model Coefficient of Determination (R2) • The quantity R2 is called the coefficient of determination and is often used to judge the fit of a regression model

• 0  R2  1 or

0%  R2  100%

• We often refer to R2 as the amount of variability in the data explained or ed for by the regression model.

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R2 • Higher R2 values indicate a better fit of line to data • We often refer to R2 as the amount of variability in the data explained or ed for by the regression model Y

Y

X

X

Higher R2

Lower R2

• No universal cut-off or threshold R2 value to define „good fitting‟ model versus „bad fitting model

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Coef 7.250 0.69667

SE Coef 1.301 0.09501

R-Sq = 71.0%


T 5.57 7.33

P 0.000 0.000

R-Sq(adj) = 69.6% MS 364.01 6.77

F 53.76

P 0.000

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Regression in Excel

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Significance of Regression An important part of assessing the adequacy of a linear regression model is testing statistical hypotheses about the model parameters. We can form a CI on both the slope, β1, and the intercept, β0 The most important CI is on the slope, β1, since this CI tests the significance of regression (i.e. Does the independent variable have a significant effect on the dependent variable?). A slope of 0 indicates there is no linear relationship between X and Y.

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Significance of Regression

Figure 11-5 1 = 0 indicating no linear relationship between X and Y.

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Significance of Regression

Figure 11-6 1 ≠ 0 indicating a linear relationship

between X and Y.

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Significance of Regression H0: β1 = 0 H1: β1 ≠ 0 In Excel • given CI for slope. Believe β1 ≠ 0 (i.e. X DOES have significant effect on Y) if 0 not in CI. • given t test statistic for slope. Reject H0 (i.e. X DOES have significant effect on Y) if |T| > tα/2,n-2 or if P-value < α In Minitab • given t test statistic for slope. Reject H0 (i.e. X DOES have significant effect on Y) if |T| > tα/2,n-2 or if P-value < α

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Regression in Excel

β1≠ 0? : when β1≠ 0 then X significantly affects Y Use CI - does CI contain 0? If CI does not contain 0, then we believe X significantly affects Y

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Coef 7.250 0.69667

SE Coef 1.301 0.09501

R-Sq = 71.0%


T 5.57 7.33

P 0.000 0.000

R-Sq(adj) = 69.6% MS 364.01 6.77

F 53.76

P 0.000

Hardwood concentration DOES have a significant effect on bag strength

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Another Example • How does weight (in lbs) affect time to run 100 yards? weight run time • dependent (Y)? • Independent (X)?

120 142 137 129 137 112 148 132 126

16 37 27 22 31 12 39 26 17

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Plot data in Excel • Straight line good fit? • Do Excel Scatter plot 45 40

run time (sec)

35 30 25 20 15 10 5 0 80

100

120

140

160

weight (lbs)

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Find Regression Equation • By Hand • In Excel > Tools

> Data Analysis >Regression

• In Minitab

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Excel Regression Output SUMMARY OUTPUT Regression Statistics Multiple R 0.969596224 R Square 0.940116838 Adjusted R Square 0.931562101 Standard Error 2.453310621 Observations 9

Line a good fit? R2=0.94

ANOVA df Regression Residual Total

Intercept weight

1 7 8

SS MS F Significance F 661.4244245 661.4244 109.8943 1.5663E-05 42.13113102 6.018733 703.5555556

Coefficients Standard Error t Stat P-value Lower 95% Upper 95% -82.0970885 10.2700248 -7.993855 9.16E-05 -106.38184 -57.81234 0.816461366 0.077883967 10.48305 1.57E-05 0.63229505 1.0006277

Time = -82.1 + 0.82(weight)

95% CI on β1

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Excel Regression Output • Is this line a good fit? R2=0.94 • Does weight significantly affect run time? Is weight useful in predicting run time? • 95% CI on β1 = (0.63, 1.00) • Since CI does not contain 0, we believe β1 ≠ 0, therefore weight does significantly affect run time.

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Predictions • Predict run time for weight of 135 lbs.

• Predict run time for weight of 50 lbs.

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Predictions • Predictions only meaningful in general range of original data. • Be very careful extrapolating regression model Y

IE 563 Dr. G.W. DePuy, UofL

X

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Another Example • How does outdoor temperature affect time (in min) to unload truck? obs # temp unload time 7 6 12 2 13 5 16 11 1 9 4 14 3 10 18 15 17 8

• dependent (Y)? • Independent (X)?

40 40 40 50 50 50 60 60 60 70 70 70 80 80 80 90 90 90

37 29 35 39 40 44 47 44 46 49 53 55 46 47 43 36 37 34

469

Truck Unload Example Regression Analysis: unload versus temp The regression equation is unload = 36.8 + 0.0848 temp Predictor Constant temp S = 6.94574

Coef 36.768 0.08476

SE Coef 6.442 0.09586

R-Sq = 4.7%

T 5.71 0.88

P 0.000 0.390

R-Sq(adj) = 0.0%

Does outdoor temperature affect time (in min) to unload truck? IE 563 Dr. G.W. DePuy, UofL Good regression model for this data? Why?

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Truck Unload Example • Plot data – straight line a good fit? Scatterplot of unload vs temp 55 50

unload

45

40

35

30 40

50

60

70

80

90

temp


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Truck Unload Example • Include higher order in model e.g. temp2 term Unload time = β0 + β1temp + β2temp2 • Now regression model has more than one term – called Multiple Linear Regression

93

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Multiple Linear Regression • General multiple linear regression model with k regressors Y = β0 + β1X1 + β2X2 + … + βkXk + ε • For our truck unload example • X1 = temp • X2 = temp2 • Model a linear function of parameters β0, β1, β2, …, βk • Testing for significance of regression now becomes: H0: β1 = β2 = … = βk = 0 H1: βk ≠0 for at least one k

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Multiple Linear Regression • Do not add too many to model • Adding will always make R2 increase • If we had 8 data points, a model with 7 higher order will fit perfectly, i.e. R2 = 100% Y = β0 + β1X + β2X2 + β3X3 + … + β7X7

• But too many will be difficult to interpret and will not give good predictions • So keep model as low order as possible

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Multiple Linear Regression • For Multiple Linear Regression use Adjusted R2 to measure model adequacy • Adjusted R2 will not automatically increase as added to model. • R2 adjusted by the number of in model

 n 1  2 1  R 2  Radj  1   n p • Where n = # data points and p = # in model (including constant)

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Truck Unload Example • Include temp2 term • Add temp2 column to Minitab worksheet Calc Calculator


95

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Truck Unload Example


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Truck Unload Example • Include temp2 term in Regression model


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Truck Unload Example Regression Analysis: unload versus temp, temp^2 The regression equation is unload = - 55.7 + 3.14 temp - 0.0235 temp^2 Predictor Constant temp temp^2

Coef -55.71 3.1413 -0.023512

S = 3.19108

SE Coef 12.22 0.3945 0.003015

R-Sq = 81.1%

T -4.56 7.96 -7.80

P 0.000 0.000 0.000

R-Sq(adj) = 78.6%

Analysis of Variance Source Regression Residual Error Total

DF 2 15 17

SS 656.87 152.75 809.61

MS 328.43 10.18

F 32.25

Model a good fit to data? Does temperature affect truck unload time?

P 0.000

H0:β1=β2=0

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Truck Unload Example • Predict unload time for temperature of 55 Unload time = -55.7 + 3.14(55) – 0.0235(552) = 45.94 min • 95% PI on unload time for temperature of 55 • In Minitab, „Options‟ menu of Regression Menu, input values for each model term • temp = 55 • temp2 = 552 = 3025

97

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Truck Unload Example Predicted Values for New Observations New Obs

Fit

SE Fit

1

45.937

1.046

95% CI

95% PI

(43.708, 48.166)

(38.779, 53.094)

Values of Predictors for New Observations

New Obs

temp

temp^2

1

55.0

3025

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Truck Unload Example in Excel • Columns for temp, temp2, unload time • Include columns for both temp & temp2

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Truck Unload Example in Excel SUMMARY OUTPUT

Model a good fit to data?

Regression Statistics Multiple R 0.90074139 R Square 0.811335052 Adjusted R Square 0.786179726 Standard Error 3.191083809 Observations 18

Does temperature affect truck unload time?


Intercept temp temp^2

483

2 15 17

SS MS F Significance F 656.865873 328.4329365 32.25301233 3.69559E-06 152.7452381 10.18301587 809.6111111

Coefficients Standard Error t Stat P-value Lower 95% -55.7119048 12.22389608 -4.55762258 0.000377434 -81.7665224 3.141309524 0.394454127 7.96368782 9.10675E-07 2.300550459 -0.0235119 0.003015291 -7.79755802 1.179E-06 -0.02993884

Upper 95% -29.6572871 3.982068589 -0.01708496

Multiple Linear Regression Another example • A pastry chef is interested in how number of eggs and amount of milk effect cake height. • Dependent (Y)? • Independent (X)?

eggs 1 1 1 2 2 2 3 3 3 1 1 1 2 2 2 3 3 3

milk 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1 1 1 1 1 1 1 1 1

height 2.3 2.1 2.5 3.4 3.3 3 4.2 3.9 4.3 2.4 2.7 2.3 2.8 2.9 2.5 2.9 3 3.2

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Multiple Linear Regression • Include for Eggs, Milk, and Eggs*Milk • Columns for all and response in Minitab

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Multiple Linear Regression Regression Analysis: height versus eggs, milk, eggs*milk The regression equation is height = 0.600 + 1.55 eggs + 1.58 milk - 1.27 eggs*milk

Predictor Constant eggs milk eggs*milk

Coef 0.6000 1.5500 1.5778 -1.2667

S = 0.184089

SE Coef 0.3630 0.1680 0.4592 0.2126

R-Sq = 93.2%

T 1.65 9.22 3.44 -5.96

P 0.121 0.000 0.004 0.000

What significantly effect cake height?

R-Sq(adj) = 91.8%

Is this model a good fit?

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Multiple Linear Regression • Predict cake height for 1.5 eggs and ¾ cup of milk. • height = 0.600 + 1.55 eggs + 1.58 milk 1.27 eggs*milk • height = 0.600 + 1.55 *(1.5) + 1.58 *(.75) 1.27 *(1.5)*(.75) = 2.6833”

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Multiple Linear Regression • Interaction of #eggs and milk Interaction Plot for height Fitted Means

milk 0.5 1.0

4.0

Mean

3.5

3.0

2.5

1

2 eggs

3

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Multiple Linear Regression in Excel • Columns for eggs, milk, eggs*milk, height • Include columns for eggs, milk, eggs*milk

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Multiple Linear Regression in Excel SUMMARY OUTPUT

Is this model a good fit?

Regression Statistics Multiple R 0.96564146 R Square 0.93246342 Adjusted R Square 0.9179913 Standard Error 0.18408935 Observations 18

What significantly effect cake height?


Intercept eggs milk eggs*milk

3 14 17

SS MS 6.550555556 2.18351852 0.474444444 0.03388889 7.025

Coefficients Standard Error 0.6 0.363029095 1.55 0.168049816 1.57777778 0.459199518 -1.26666667 0.212568072

F Significance F 64.431694 1.9533E-08

t Stat P-value Lower 95% Upper 95% 1.65276009 0.12062076 -0.17861997 1.37861997 9.22345549 2.5181E-07 1.18956899 1.91043101 3.43593082 0.00401543 0.59289277 2.56266279 -5.95887546 3.4936E-05 -1.72257984 -0.8107535

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Test #2 here • Take home exam • You may use book, notes, calculator • You may not discuss test with anyone other than me. • Due • Covers Chapters 6, 7, 8, 9, 11

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Good-Bye! • Thanks! • Keep working hard – you will be successful!! • Good luck!!

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