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Simultaneous inference Estimating (or testing) more than one thing at a time (such as β 0 and β 1 ) and feeling confident about it …

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Simultaneous inference. Estimating (or testing) more than one thing at a time (such as β 0 and β 1 ) and feeling confident about it …. Simultaneous inference we’ll be concerned about …. Estimating β 0 and β 1 jointly. Estimating more than one mean response, E(Y), at a time. - PowerPoint PPT Presentation

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Simultaneous inference

Estimating (or testing) more than one thing at a time (such as β0 and β1) and

Simultaneous inference we’ll be concerned about …

• Estimating β0 and β1 jointly.

• Estimating more than one mean response, E(Y), at a time.

• Predicting more than one new observation at a time.

Why simultaneous inference is important

• A 95% confidence interval implies a 95% chance that the interval contains β0.

• A 95% confidence interval implies a 95% chance that the interval contains β1.

• If the intervals are independent, then have only a (0.95×0.95) ×100 = 90.25% chance that both intervals are correct.

• (Intervals not independent, but point made.)

Terminology

• Family of estimates (or tests): a set of estimates (or tests) which you want all to be simultaneously correct.

• Statement confidence level: the confidence level, as you know it, that is, for just one parameter.

• Family confidence level: the confidence level of the whole family of interval estimates (or tests).

Examples

• A 95% confidence interval for β0 – the 95% is a statement confidence level.

• A 95% confidence interval for β1 – the 95% is a statement confidence level.

• Consider family of interval estimates for β0 and β1. If a 90.25% chance that both intervals are simultaneously correct, then 90.25% is the family confidence level.

Bonferroni joint confidence intervals for β0 and β1

• GOAL: To formulate joint confidence intervals for β0 and β1 with a specified family confidence level.

• BASIC IDEA: – Make statement confidence level for β0 higher

– Make statement confidence level for β1 higher

– So that the family confidence level for (β0 , β1) is at least (1-α)×100%.

Recall: Original confidence intervals

00 2,2

1 bsntb

For β0:

11 2,2

1 bsntb

For β1:

Goal is to adjust the t-multiples so that family confidence coefficient is 1-α.

That is, we need to find the α* to put into the above formulas to achieve the desired family coefficient of 1- α.

A little derivation

• Let A1 = the event that first confidence interval does not contain β0 (i.e., incorrect).

• So A1C

= the event that first confidence interval contains β0 (i.e., correct).

• P(A1) = α and P(A1C) = 1- α

A little derivation (cont’d)

• Let A2 = the event that second confidence interval does not contain β1 (i.e., incorrect).

• So A2C

= the event that second confidence interval contains β1 (i.e., correct).

• P(A2) = α and P(A2C) = 1- α

Becoming a not so little derivation…

A1 A2

A1 or A2 A1C and A2

C

We want P(A1C and A2

C) to be at least 1-α.

P(A1C and A2

C) = 1 – P(A1 or A2) = 1 – [P(A1)+P(A2) – P(A1 and A2)]= 1 – P(A1) – P(A2) + P(A1 and A2)]≥ 1 – P(A1) – P(A2)= 1 – α – α= 1 – 2α

So, we need α* to be set to α/2.

Bonferroni joint confidence intervals

00 2,2

1 bsntb

11 2,2

1 bsntb

00 2,4

1 bsntb

11 2,4

1 bsntb

Typically, the t-multiple in this setting is called the Bonferroni multiple and is denoted by the letter B.

Example: 90% family confidence interval

The regression equation ispunt = 14.9 + 0.903 leg

Predictor Coef SE Coef T PConstant 14.91 31.37 0.48 0.644leg 0.9027 0.2101 4.30 0.001

n=13 punters t(0.975, 11) = 2.201

9.83,1.54)37.31(201.29.14:0 36.1,44.0)21.0(201.290.0:1

We are 90% confident that β0 is between -54.1 and 83.9 and β1 is between 0.44 and 1.36.

A couple of more points about Bonferroni intervals

• Bonferroni intervals are most useful when there are only a few interval estimates in the family (o.w., the intervals get too large).

• Can specify different statement confidence levels to get desired family confidence level.

• Bonferroni technique easily extends to g interval estimates. Set statement confidence levels at 1-(α/g), so need to look up 1- (α/2g).

Bonferroni intervals for more than one mean response at a time

To estimate the mean response E(Yh) for g different Xh values with family confidence coefficient 1-α:

hh YsBY ˆˆ

where:

2,

21 n

gtB

g is the number of confidence intervals in the family

Example: Mean punting distance for leg strengths of 140, 150, 160 lbs.

Predicted Values for New Observations

New Fit SE Fit 95.0% CI 95.0% PI

140 141.28 4.88 (130.55,152.01) (103.23,179.33) 150 150.31 4.63 (140.13,160.49) (112.41,188.20) 160 159.33 5.28 (147.72,170.95) (121.03,197.64)

n=13 punters t(0.99, 11) = 2.718

5.154,0.12888.4718.228.141 9.162,7.13763.4718.231.150

7.173,0.14528.5718.233.159

We are 94% confident that the mean responses for leg strengths of 140, 150, 160 pounds are …

Two procedures for predicting g new observations simultaneously

• Bonferroni procedure

• Scheffé procedure

• Use the procedure that gives the narrower prediction limits.

Bonferroni intervals for predicting more than one new obs’n at a time

To predict g new observations Yh for g different Xh values with family confidence coefficient 1-α:

predsBYh ˆ

where:

2,

21 n

gtB

g is the number of prediction intervals in the family

222 ˆ)( SEFitMSEYsMSEpreds h

Scheffé intervals for predicting more than one new obs’n at a time

To predict g new observations Yh for g different Xh values with family confidence coefficient 1-α:

predsSYh ˆ

where:

2,;12 ngFgS

g is the number of prediction intervals in the family

222 ˆ)( SEFitMSEYsMSEpreds h

Example: Punting distance for leg strengths of 140 and 150 lbs.

n = 13 punters

Bonferroni multiple: 201.211,975.0213,22

10.01

ttB

Suppose we want a 90% family confidence level.

Scheffé multiple: 39.286.22)11,2;10.01(2 FS

Since B is smaller than S, the Bonferroni prediction intervals will be narrower … so use them here instead of the Scheffé intervals.

Example: Punting distance for leg strengths of 140 and 150 lbs.

Predicted Values for New Observations

New Fit SE Fit 95.0% CI 95.0% PI

140 141.28 4.88 (130.55,152.01) (103.23,179.33) 150 150.31 4.63 (140.13,160.49) (112.41,188.20)

n=13 punters s(pred(140)) = 17.28

3.179,2.10328.17201.228.141 2.188,4.11221.17201.231.150

There is a 90% chance that the punting distances for leg strengths of 140 and 150 pounds will be…

s(pred(150)) = 17.21

Simultaneous prediction in Minitab

• Stat >> Regression >> Regression …• Specify predictor and response.• Under Options …, In “Prediction intervals

for new observations” box, specify a column name containing multiple X values. Specify confidence level.

• Click on OK. Click on OK.• Results appear in session window.