sections 6.7, 6.8, 7.7 (note: the approach used here to present the material in these sections is...

Sections 6.7, 6.8, 7.7(Note: The approach used here to present the material in these sections is substantially different from the approach used in the textbook.)

Recall: If X and Y are random variables with E(X) = X , E(Y) = Y , Var(X) = X

2 , Var(Y) = Y2 , and Cov(X,Y) = XY , then the least

squares line for predicting Y from X isy = Y + Y

— (x – X)X

or y = Y – Y— X +X

Y — x X

a bThe least squares line is derived in Section 4.2 by minimizing

E{[Y – (a + bX)]2} .Consider a set of observed data (x1 , y1) , (x2 , y2) , … , (xn , yn ) . Imagine that we treat this data as describing a joint p.m.f. for two random variables X and Y where each points is assigned a probability of 1/n.

Then, we see that

plays the role of E(X) = X ,

plays the role of E(Y) = Y ,

plays the role of Var(X) = X2 ,

plays the role of Var(Y) = Y2 , and

plays the role of Cov(X,Y) = XY .

xi = 1— n

yi = 1— n

(xi – x)2 = sx2n 1

n – 1—— n

(yi – y)2 = sy2n 1

n – 1—— n

(xi – x)(yi – y) =n 1

we shall complete this equation shortly.

We define the sample covariance to be c = , and

we define the sample correlation to be r =

Consequently, the least squares line for predicting Y from X is

This least squares line minimizes

(xi – x)(yi – y)n

n – 1

c—— . sxsy

The sample correlation r is a measure of the strength and direction of a linear relationship for the sample in the same way that the correlation is a measure of the strength and direction of a linear relationship for the two random variables X and Y.

plays the role of E(X) = X ,

plays the role of E(Y) = Y ,

plays the role of Var(X) = X2 ,

plays the role of Var(Y) = Y2 , and

plays the role of Cov(X,Y) = XY .

xi = 1— n

yi = 1— n

(xi – x)2 = sx2n 1

n – 1—— n

(yi – y)2 = sy2n 1

n – 1—— n

(xi – x)(yi – y) =n 1

n – 1—— c n

We define the sample covariance to be c = , and

we define the sample correlation to be r =

Consequently, the least squares line for predicting Y from X is

This least squares line minimizes

y = y + r sy— (x – x) sx

or y = y – r sy— x + sx

syr — x sx

(xi – x)(yi – y)n

n – 1

c—— . sxsy

The sample correlation r is a measure of the strength and direction of a linear relationship for the sample in the same way that the correlation is a measure of the strength and direction of a linear relationship for the two random variables X and Y.

[yi – (a + bxi)]2 .i = 1

r = +1 r close to +1 r is positive

r = –1 r close to –1 r is negative

r close to 0 r close to 0 r is negativer close to 0

Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi . Then the joint p.d.f. of Y1 , Y2 , … , Yn is

exp—————————— =

[yi – (0 + 1xi)]2

– ——————— 22

exp————————————

n (2)n/2

[yi – (0 + 1xi)]2

– ———————— 22

for – < y1 < , – < y2 < , … , – < yn <

If we treat this joint p.d.f. as a function L(0 , 1), that is, a function of the unknown parameters 0 and 1 , then we can find the maximum likelihood estimates for 0 and 1 by maximizing the function L(0 , 1).It is clear the function L(0 , 1) will be maximized when

is minimized.[yi – (0 + 1xi)]2i = 1

The previous result concerning the least squares line for predicting Y from X with a sample of data points tells us that the

mle of 1 is 1 =

and the

mle of 0 is 0 =

SyR — = sx

(xi – x)(Yi – Y)n

(xi – x)2n

(xi – x)Yi

(xi – x)2n

(xj – x)2n

(xi – x)Yi

Y – RSy— x =sx

n Yi— n

(xj – x)2n

(xi – x)Yi =

n 1— n

(xj – x)2n

(xi – x)Yi

Suppose we are interested in predicting a person's height from the person's length of stride (distance between footprints). The following data is recorded for a random sample of 5 people:

Length of Stride (inches) 14 13 21 25 17Height (inches) 61 54 63 72 59

Find the equation of the least squares line for predicting a person's height from the person's length of stride.

The slope of the least squares line is120—— = 1.2 .100

The intercept of the least squares line is 61.8 – (1.2)(18) = 40.2 .

The least squares line can be written y = 40.2 + 1.2x .

(b) Suppose we assume that the height of humans has a normal distribution with mean 0 + 1x and variance 2, where x is the length of stride. Find the maximum likelihood estimators for 0 and 1 .

The mle of 1 is120—— = 1.2 .100

The mle of 0 is 61.8 – (1.2)(18) = 40.2 .

(a) Use Theorem 5.5-1 (Class Exercise 5.5-1) to find the distribution of the maximum likelihood estimator of 1.

Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi .

(xj – x)2n

(xi – x)Yi

has a normal distribution with

mean n

(xj – x)2n

(xi – x)(0 + 1xi) = n

(xj – x)2n

(xi – x)[0 + 1x + 1(xi – x)] =

(xj – x)2n

(xi – x)[0 + 1x] + n

(xj – x)2n

(xi – x) 1(xi – x) = 1 n

(xj – x)2n

(xi – x)2

and variance n

(xj – x)2n

(xi – x)2 =

(xj – x)2n

(xi – x)2

2. - continued

(b) Use Theorem 5.5-1 (Class Exercise 5.5-1) to find the distribution of the maximum likelihood estimator of 0.

n 1— n

(xj – x)2n

(xi – x)Yi

xhas a normal distribution with

mean i = 1

n 1— n

(xj – x)2n

(xi – x) x(0 + 1xi) =

n 1— n

(xj – x)2n

(xi – x) x(0 + 1xi) =

n(0 + 1xi) –

0 + 1x – x n

(xj – x)2n

(xi – x)(0 + 1xi) =

We already found this in part (a).

0 + 1x – 1x = 0

and variance i = 1

n 1— n

(xj – x)2n

(xi – x) x 2

n 1— n2

(xj – x)2n

2 (xi – x) x

n j = 1

(xj – x)2n

(xi – x)2

2 1— n

(xi – x)2

Suppose we treat the joint p.d.f. as a function L(0 , 1 , 2), that is, a function of the three unknown parameters (instead of just two). Then, analogous to Text Example 6.1-3, we find that the maximum likelihood estimates for 0 and 1 are the same as previously derived, and that the maximum likelihood estimator for 2 is

[Yi – (0 + 1xi)]2i = 1

Recall: If Y1 , Y2 , … , Yn are independent with each having a N(, 2) distribution (i.e., a random sample from a N(, 2) distribution), then

Y = has a distribution,

has a distribution, and

n N( , )2

(n – 1)S2

———– 2

2(n – 1)

the random variables Y and are(n – 1)S2

———– 2 independent.

Analogous results for the more general situation previously considered can be proven using matrix algebra.

Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi . Then

1 has a distribution,

0 has a distribution,

^ N( 1 , )2

(xi – x)2

^ N( 0 , )

2 1— n

(xi – x)2

has a distribution,

random variables and are independent, and

random variables and are independent.

[Yi – (0 + 1xi)]2i = 1

2(n – 2)

[Yi – (0 + 1xi)]2i = 1

(Yi – Y)2 =n(Yi – Y)2 +

n ^(Yi – Yi)2

This is called the total sum of squares and is denoted SST.

This is called the regression sum of squares and is denoted SSR.

This is called the error (residual) sum of squares and is denoted SSE.

Since, as we have noted, SSE / 2 has a 2(n – 2) distribution, we say that the df (degrees of freedom) associated with SSE is n – 2 .

If Y1 , Y2 , … , Yn all have the same mean, that is, if 1 = 0, then SST / 2 has a 2(n – 1) distribution; consequently, the df associated with SST is n – 1 .

If Y1 , Y2 , … , Yn all have the same mean, that is, if 1 = 0, then it can be shown that SSR and SSE are independent, and that SSR / 2 has a 2(1) distribution; consequently, the df associated with SSR is 1 .

For each i = 1, 2, …, n, we define the random variable Yi = 0 + 1xi , that is, Yi is the predicted value corresponding to xi . With appropriate algebra, it can be shown that

3. Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi . Prove that SST = SSR + SSE .

First, we observe that for i = 1, 2, …, n, Yi = 0 + 1xi = ^ ^^ Y + 1(xi – x)^

SST = (Yi – Y)2 =n

ni = 1

1(xi – x) + (Yi – Y) – 1(xi – x) =^ ^2

ni = 1

[1(xi – x)]2 +^ [(Yi – Y) – 1(xi – x)]2 +^

21(xi – x)[(Yi – Y) – 1(xi – x)] =^ ^

ni = 1

^[1(xi – x)]2 + ni = 1

[Yi – Y – 1(xi – x)]2 +^

2 1(xi – x)(Yi – Y) – 12(xi – x)2 =^ ^n

ni = 1

(Yi – Y)2 +i = 1

n ^(Yi – Yi)2 +

2 1 (xi – x)(Yi – Y) – 12 (xi – x)2 =^ ^n

ni = 1

SSR + SSE +

(xi – x)2n

2 (xi – x)(Yi – Y) –ni = 1

(xi – x)2n

(xi – x)2ni = 1

= SSR + SSE

A mean square is a sum of squares divided by its degrees of freedom.

The error (residual) mean square is MSE = . SSE——n – 2

The regression mean square is MSR = . SSR—— 1

yTotal variation in Y is based on SST.

Variation in Y explained by the linear relationship with X is based on SSR.

Variation in Y explained by random error is based on SSE.

If Y1 , Y2 , … , Yn all have the same mean, that is, if 1 = 0, then

E(SSR / 2) = and E

If 1 0, then E(SSR / 2) =

SSE / 2

——— = n – 2

(Yi – Y)2i = 1

n ^ [1(xi – x)]2^

E ————— =2

E —————– =2

(xi – x)2i = 1

———— 2 E(1

2) =^(xi – x)2

———— 2 Var(1) + [E(1)]2 =^ ^

1 + 12 > 1 .

(xi – x)2i = 1

———— 2

This suggests that if 1 = 0, then the ratio of SSR / 2 to is expected to be close to one, but if 1 0, then this ratio is expected to be

SSE / 2

——— n – 2

larger than one.

If 1 = 0, then since SSR and SSE are independent, then the ratio of

SSR / 2 to must have an distribution.SSE / 2

——— n – 2

f( , )1 n – 2

Consequently, we can perform the hypothesis test

H0: 1 = 0 vs. H1: 1 0 .

by using the test statistic F =

We reject H0 (in favor of H1) when f > f(1, n – 2) .

SSR / 2 ———— =SSE / 2

——— n – 2

MSR——– . MSE

The linear relationship (correlation) between X and Y is not significant.

The linear relationship (correlation) between X and Y is significant.

The calculations in a regression leading up to the f statistic are often organized into an analysis of variance (ANOVA) table such as the following:

Source df SS MS f p-value

Regression

It can be shown that the squared sample correlation is

which is often called the proportion of variation in Y explained by X. The standard error of estimate is defined to be

SSR1 MSR

SSEn – 2 MSE

n – 1

MSR / MSE

SSR—— . SST

(This is done in Class Exercise #7.)

1. - continued

(c) Find the sums of squares SSR, SSE, and SST; then construct the ANOVA table and perform the corresponding f test with = 0.05, find and interpret the squared sample correlation, and find the standard error of estimate.

SSR = (Yi – Y)2 =i = 1

n ^ ni = 1

^ (xi – x)2 =12 (1.2)2(100) = 144

SST = (Yi – Y)2 =i = 1

n 174.8

SSE = SST – SSR = 174.8 – 144 = 30.8

Regression

1441 144

30.83 10.267

0.025< p < 0.05

Since f = 14.03 > f0.05(1,3) = 10.13, we reject H0. We conclude that the slope in the linear regression of height on stride length is different from zero (0.025< p-value < 0.05), and the results suggest that this slope is positive.(Note: We could alternatively conclude that the linear relationship or correlation is significant, and that the results suggest a positive linear relationship or correlation.)

R2 = 144 / 174.8 = 82.4% of the variation in height is explained by stride length.

The standard error of estimate is SXY = 10.267 = 3.20 inches.

The prediction of grip strength from age for right-handed males is of interest. It is assumed that for any age x in years, where 10 x 25, Y = “grip strength in pounds” has a N(x , 2) distribution where x = 0 + 1x . For a random sample of right-handed males, the following data are recorded:

Age (years) 15 17 19 11 16 22 17 25 12 14 25 23Grip Strength (lbs.) 50 54 66 46 58 54 64 80 46 70 76 80

Obtain the calculations below from a calculator and from SPSS.

x =n = y =

(xi – x)2 =i = 1

n (yi – y)2 =i = 1

(xi – x)(yi – y) =ni = 1

12 18 62

256 1728

r =512 + 0.770

Descriptive Statistics

62.00 12.534 12

18.00 4.824 12

Mean Std. Deviation N

Model Summaryb

.770a .593 .552 8.390Model1

R R SquareAdjustedR Square

Std. Error ofthe Estimate

Predictors: (Constant), agea.

Dependent Variable: gripb.

Correlations

1.000 .770

.770 1.000

. .002

.002 .

Pearson Correlation

Sig. (1-tailed)

grip age

8. - continued

Find the equation of the least squares line from a calculator and from SPSS.

The least squares line can be written y = 26 + 2x .

Write a one-sentence interpretation of the slope in the least squares line and a one-sentence interpretation of the intercept in the least squares line.

Grip strength appears to increase on average by about 2 pounds with each increase of one year in age.

The intercept is the mean grip strength at age zero, which makes no sense in this situation.

Find r2, and write a one-sentence interpretation. (d)

(e) Find the standard error of estimate.

r2 = 0.593

About 59.3% of the variation in grip strength is explained by age.

s = 8.390

8. - continued

(f) Construct the ANOVA table and perform the corresponding f test with = 0.05.

SSR = (Yi – Y)2 =i = 1

n ^ ni = 1

^ (xi – x)2 =12 (2)2(256) = 1024

SST = (Yi – Y)2 =i = 1

n 1728

SSE = SST – SSR = 1728 – 1024 = 704

Regression

10241 1024

70410 70.4

p < 0.01

The test statistic is f = 14.55

The critical region with = 0.05 is f 4.96 .

4.96 = f0.05(1,10)

ANOVAb

1024.000 1 1024.000 14.545 .003a

704.000 10 70.400

1728.000 11

Regression

Residual

Model1

Sum ofSquares df Mean Square F Sig.

Predictors: (Constant), agea.

Dependent Variable: gripb.

Since f = 14.55 > f0.05(1,10) = 4.96, we reject H0. We conclude that the slope in the linear regression of grip strength on age is different from zero (p < 0.01), and the results suggest that this slope is positive.(Note: We could alternatively conclude that the linear relationship or correlation is significant, and that the results suggest a positive linear relationship or correlation.)

The p-value issmaller than 0.01 (from the table) or p = 0.003 (from the SPSS output).

4. Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi . Show that the maximum likelihood estimator of 2 is not unbiased; then, find a constant multiple of this maximum likelihood estimator which is an unbiased estimator of 2.

[Yi – (0 + 1xi)]2i = 1

E = E =[Yi – (0 + 1xi)]2

— (n – 2) n

An unbiased estimator of 2 is n——n – 2

[Yi – (0 + 1xi)]2i = 1

(Yi – Yi)2i = 1

————— n – 2

= MSE SSE—— =n – 2

1^ – 1

(xi – x)2

SSE / 2

——— n – 2

=1^ – 1

(xi – x)2

has a distribution.t(n – 2)

0^ – 0

1— n

(xi – x)2

SSE / 2

——— n – 2

=0^ – 0

1— n

(xi – x)2

5. Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n,

i = 0 + 1xi .

For a given value x0 , use Theorem 5.5-1 (Class Exercise 5.5-1) to

find the distribution of Y | x0 = 0 + 1x0 (the predicted value of Y

corresponding to the value x0).

Y | x0 = 0 + 1x0 =^ ^^ Y + 1(x0 – x) =^

n Yi— n

(xj – x)2ni = 1

(xi – x)Yi =

(x0 – x)

n 1— n

(xj – x)2n

(xi – x)Yi

(x0 – x)has a normal distribution with

mean E(0 + 1x0) =

and variance i = 1

n 1— n

(xj – x)2n

(xi – x) (x0 – x)2

We can find this with algebra analogous to that in Class Exercise #2(b).

2 1— n

(xi – x)2n

(x0 – x)2

^ ^ 0 + 1x0

In Class Exercise #2(b), this was a minus sign. In Class Exercise #2(b), this was just x .

For a given value x0 , we define Y | x0 = 0 + 1x0 to be the predicted value of Y corresponding to the value x0 ; this predicted value has a

N( , )0 + 1x0 2 1— n

(xi – x)2n

(x0 – x)2

random variables and are independent.

[Yi – (0 + 1xi)]2i = 1

Y | x0 = 0 + 1x0^ ^^

1— n

(xi – x)2

SSE / 2

——— n – 2

1— n

(xi – x)2

(x0 – x)2 (x0 – x)2

0 + 1x0^ ^ (0 + 1x0) –0 + 1x0

^ ^ (0 + 1x0)

We assume min{x1 , x2 , … , xn} x0 max{x1 , x2 , … , xn}, since prediction outside the range of the data may not be valid.

distribution, and

Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi .

Derive a 100(1 – )% confidence interval for the slope 1 .

1^ – 1

(xi – x)2

t/2(n – 2) = 1 –

P – t/2(n – 2)

P 1 – t/2(n – 2) 1 1 + t/2(n – 2)

(xi – x)2

= 1 –

If 1(0) is a hypothesized value for 1 , then derive the test statistic and rejection regions corresponding to the one sided and two sided hypothesis tests for testing H0: 1 = 1(0) with significance level .

The test statistic is T =1^ – 1(0)

(xi – x)2

For H1: 1 < 1(0) , the rejection region is t – t(n – 2) .

For H1: 1 > 1(0) , the rejection region is t t(n – 2) .

For H1: 1 1(0) , the rejection region is |t| t/2(n – 2) .

8. - continued

(g) Perform the t test for H0: 1 = 0.8 vs. H1: 1 0.8 with = 0.05.

The test statistic is t =

The two-sided critical region with = 0.05 is |t| 2.228 .

2.228 = t0.025(10)

1^ – 0.8

(xi – x)2

=2 – 0.8

= 2.288 .

– 2.228The p-value isbetween 0.02 and 0.05 (from the table).

Since t = 2.288 > t0.025(10) = 2.228, we reject H0. We conclude that the slope in the linear regression of grip strength on age is different from 0.8 lbs. (0.02 < p < 0.05), and the results suggest that this slope is greater than 0.8 lbs.(Note: We could alternatively conclude that the average change in grip strength is different from 0.8 lbs. per year, and that this change is greater than 0.8 lbs. per year.)

8. - continued

(h) Considering the results of the hypothesis tests in parts (f) and (g), explain why a 95% confidence interval for the slope in the regression would be of interest. Then find and interpret the confidence interval.

1 – t/2(n – 2) 1 1 + t/2(n – 2)^MSE

(xi – x)2

Since rejecting H0 in part (f) suggests that the hypothesized zero slope is not correct, and rejecting H0 in part (g) suggests that the hypothesized slope of 0.8 is not correct, a 95% confidence interval will provide us with some information about the value of the slope, which estimates the average change in grip strength with an increase of one year in age.

2 (2.228)

70.4—— 256

0.832 and 3.168

We are 95% confident that the slope in the regression to predict grip strength from age is between 0.832 and 3.168 lbs.

6. - continued

(c) Derive a 100(1 – )% confidence interval for the intercept 0 .

0^ – 0

1— n

(xi – x)2

t/2(n – 2) = 1 –

P – t/2(n – 2)

P 0 – t/2(n – 2)^ 1— n

(xi – x)2

0 + t/2(n – 2)^ 1— n

(xi – x)2

= 1 –

If 0(0) is a hypothesized value for 0 , then derive the test statistic and rejection regions corresponding to the one sided and two sided hypothesis tests for testing H0: 0 = 0(0) with significance level .

The test statistic is T =

For H1: 0 < 0(0) , the rejection region is t – t(n – 2) .

For H1: 0 > 0(0) , the rejection region is t t(n – 2) .

For H1: 0 0(0) , the rejection region is |t| t/2(n – 2) .

0^ – 0(0)

1— n

(xi – x)2

8. - continued

(i) Perform the t test for H0: 0 = 0 vs. H1: 0 0 with = 0.05.

The test statistic is t =0^ – 0

1— n

(xi – x)2

26 – 0

1—12

—— 256

The two-sided critical region with = 0.05 is|t| 2.228 . 2.228 =

t0.025(10) – 2.228

The p-value isbetween 0.02 and 0.05 (from the table) or p = 0.024 (from the SPSS output).

Since t = 2.668 > t0.025(10) = 2.228, we reject H0. We conclude that the intercept in the linear regression of grip strength on age is different from zero (0.02 < p < 0.05), and the results suggest that this intercept is positive.

8. - continued

(j) Considering the results of the hypothesis test in part (i), explain why a 95% confidence interval for the intercept in the regression would be of interest. Then find and interpret the confidence interval.

Since rejecting H0 in part (i) suggests that the hypothesized zero intercept is not correct, a 95% confidence interval will provide us with some information about the value of the intercept.

0 – t/2(n – 2)^ 1— n

(xi – x)2

0 + t/2(n – 2)^ 1— n

(xi – x)2

26 (2.228)

1—12

+70.4 182

—— 256

4.288 and 47.712

We are 95% confident that the intercept in the regression to predict grip strength from age is between 4.288 and 47.712 lbs.

6. - continued

(e) For a given value x0 , we can call E(Y | x0) = 0 + 1x0 the mean of

Y corresponding to the value x0 , and an unbiased estimator of

this mean is Y | x0 = 0 + 1x0 (from Class Exercise #5). Derive

a 100(1 – )% confidence interval for E(Y | x0) = 0 + 1x0 .

t/2(n – 2) = 1 –

P – t/2(n – 2) 1— n

(xi – x)2

(x0 – x)2

–0 + 1x0^ ^ (0 + 1x0)

P 0 + 1x0 – t/2(n – 2)^

0 + 1x0 + t/2(n – 2)^

0 + 1x0

= 1 –

1— n

(xi – x)2

(x0 – x)2

1— n

(xi – x)2

(x0 – x)2

6. - continued

(f) If 0 is a hypothesized value for E(Y | x0) = 0 + 1x0 , then derive the test statistic and rejection regions corresponding to the one sided and two sided hypothesis tests for testing H0: E(Y | x0) = 0 with significance level .

The test statistic is T =

For H1: E(Y | x0) < 0 , the rejection region is t – t(n – 2) .

For H1: E(Y | x0) > 0 , the rejection region is t t(n – 2) .

For H1: E(Y | x0) 0 , the rejection region is |t| t/2(n – 2) .

1— n

(xi – x)2

(x0 – x)2

–0 + 1x0^ ^ 0

8. - continued

(k) Perform the t test forH0:

vs.H1:

with = 0.05, that is, H0: E(Y | x = 20) = 80 vs. H1: E(Y | x = 20) 80

The mean grip strength for 20 year old right-handed males is 80 lbs.

The mean grip strength for 20 year old right-handed males is different from 80 lbs.

The test statistic is t =

1— n

(xi – x)2

(x0 – x)2

–0 + 1x0^ ^ 0

=66 – 80

1—12

=(20 – 18)2

———— 256

– 5.304

The two-sided critical region with = 0.05 is|t| 2.228 . 2.228 =

t0.025(10) – 2.228

The p-value is less than 0.01 (from the table).

Since |t| = 5.304 > t0.025(10) = 2.228, we reject H0. We conclude that the mean grip strength for 20 year old right-handed males is different from 80 lbs. (p < 0.01), and the results suggest that this mean is less than 80 lbs.

8. - continued

(l) Considering the results of the hypothesis test in part (k), explain why a 95% confidence interval for the mean grip strength for 20 year old right-handed males would be of interest. Then find and interpret the confidence interval.

Since rejecting H0 in part (k) suggests that the mean grip strength for 20 year old right-handed males is not 80 lbs., a 95% confidence interval will provide us with some information about this mean.

0 + 1x0 – t/2(n – 2)^

0 + 1x0 + t/2(n – 2)^

0 + 1x0

1— n

(xi – x)2

(x0 – x)2

1— n

(xi – x)2

(x0 – x)2

66 (2.228)

60.12 and 71.88

We are 95% confident that the mean grip strength for 20 year old right-handed males is between 60.12 and 71.88 lbs.

1—12

+70.4(20 – 18)2

———— 256

Return to #6, part (g)

For a given value x0 , let Y0 be a random variable independent of Y1 , Y2 , … , Yn and having a N(0 , 2) where 0 = 0 + 1x0 , that is, Y0 is a “new” random observation. Use Theorem 5.5-1 (Class Exercise 5.5-1) to derive a 100(1 – )% prediction interval for the value of Y0 . ^ ^Y0 – (0 + 1x0) has a normal distribution with

mean 0 + 1x0 – (0 + 1x0) = 0and variance 2 + .2

1— n

+ (xi – x)2n

(x0 – x)2

11 + — n

(xi – x)2

SSE / 2

——— n – 2

(xi – x)2

^ ^Y0 – (0 + 1x0) ^ ^Y0 – (0 + 1x0)

(x0 – x)2

(x0 – x)2 11 + — n

6. (g) - continued

(xi – x)2

^ ^Y0 – (0 + 1x0)

(x0 – x)2 11 + — n

– t/2(n – 2) t/2(n – 2)

= 1 –

P (0 + 1x0) – t/2(n – 2) +

(xi – x)2

(x0 – x)2 11 + — n

(0 + 1x0) + t/2(n – 2) +

(xi – x)2

(x0 – x)2 11 + — n

= 1 –

8. - continued

(m) Find and interpret a 95% prediction interval for the grip strength for a 20 year old right-handed male.

(0 + 1x0) – t/2(n – 2) +

(xi – x)2

(x0 – x)2 11 + — n

(0 + 1x0) + t/2(n – 2) +

(xi – x)2

(x0 – x)2 11 + — n

66 (2.228)

11 + — 12

+70.4(20 – 18)2

———— 256

46.40 and 85.60

We are 95% confident that the grip strength for a randomly selected 20-year old right-handed male will be between 46.40 and 85.60 lbs.

At least 95% of 20-year old right-handed males have a grip strength between 46.40 and 85.60 lbs.

(n) For what age group of right-handed males will the confidence interval for mean grip strength and the prediction interval for a particular grip strength both have the smallest length?

18 year olds

6. - continued

(h) Consider the two sided hypothesis test H0: 1 = 0 vs. H1: 1 0 with significance level , which is one of the hypothesis tests defined in part (b). Prove that the square of the t test statistic for this hypothesis test is equal to the f test statistic in the one-way ANOVA.

1^ – 0

(xi – x)2

= MSR——– MSE

We see this from the derivation in Class Exercise #3.

7. Suppose Y1 , Y2 , … , Yn are independent with respective N(1 , 2), N(2 , 2), … , N(n , 2) distributions. Let x1 , x2 , … , xn be fixed values not all equal, and suppose that for i = 1, 2, …, n, i = 0 + 1xi . Prove that

R2 = SSR—— . SST

R2 = = Sy

R2 — — = sx

(xi – x)2

(Yi – Y)2

SSR—— SST

8. - continued

(o) Use SPSS to graph the least squares line on a scatter plot, and comment on how appropriate the linear model seems to be.

The linear model appears to be reasonable.

9. Voters in a state are surveyed. Each respondent is assigned an identification number (ID), and the following information about each is recorded: sex, area of residence (RES), political party affiliation (POL), number of children (CHL), age, yearly income (INC), job satisfaction score (JSS) where 0 = totally dissatisfied and 10 = totally satisfied, weekly hours spent watching television (TVH), and weekly hours spent listening to radio (RAD). The resulting data is as follows:ID SEX RES POL CHL AGE INC JSS TVH RAD

08 M Suburban Republican 5 35 34,000 5 15 15

27 F Urban Democrat 2 20 28,000 2 20 13

34 F Suburban Independent 4 35 71,000 7 18 17

18 M Rural Independent 7 41 35,000 8 12 20

04 M Urban Republican 3 39 55,000 4 14 15

14 M Urban Democrat 3 59 75,000 1 11 18

23 F Urban Democrat 1 20 26,000 2 10 11

39 M Rural Other 4 52 30,000 9 12 30

54 F Rural Other 2 44 27,000 7 8 14

44 M Urban Republican 0 46 53,000 0 21 10

17 F Urban Republican 2 40 45,000 3 12 14

12 F Suburban Other 2 34 34,000 5 8 10

26 F Urban Republican 2 24 30,000 2 11 15

11 M Urban Democrat 4 62 78,000 1 18 11

38 M Suburban Other 3 44 68,000 7 17 12

09 F Rural Republican 6 44 29,000 9 8 27

29 F Suburban Democrat 4 38 40,000 9 9 25

13 M Rural Independent 9 47 39,000 6 8 20

24 M Urban Democrat 3 44 60,000 2 27 4

33 F Urban Democrat 1 45 49,000 3 9 12

15 F Rural Other 2 56 39,000 8 6 27

52 F Suburban Republican 0 32 33,000 5 11 15

35 M Suburban Other 1 54 65.000 3 14 14

30 F Rural Independent 3 41 25,000 7 5 25

32 F Rural Republican 2 59 41,000 8 8 27

41 F Suburban Other 3 44 44,000 3 10 18

10 M Rural Other 3 62 45,000 10 10 23

02 M Rural Democrat 8 59 39,000 8 9 24

The data is stored in the SPSS data file survey with income entered in units of thousands of dollars. The prediction of yearly income from age is of interest, and the 30 individuals selected for the data set are treated as a random sample for simple linear regression. It is assumed that for any age x in years, where 20 x , Y = “yearly income” has a N(x , 2) distribution with x = 0 + 1x .

9. - continued

(a) Use the Analyze > Regression> Linear options in SPSS to select the variable income for the Dependent slot and select the variable age for the Independent(s) section.

To have the mean and standard deviation displayed for the dependent and independent variables, click on the Statistics button, and select the Descriptives option.

To add predicted values and residuals to the data file, click on the Save button, select the Unstandardized option in the Predicted Values section, and select the Unstandardized option in the Residuals section.

Do this exercise for homework.

10. Use the fact that the random variable

SSE—— has a distribution 2

to derive a 100(1 – )% prediction interval for 2 .

2(n–2)

SSEP < —— < = 1 –

1 – /2(n–2) 2/2(n–2)

P > —— > = 1 – SSE

1—————2

1 – /2(n–2)

1————2

/2(n–2)

P > 2 > = 1 – SSE—————2

1 – /2(n–2)

SSE————2

/2(n–2)

P < 2 < = 1 – SSE—————2

1 – /2(n–2)

SSE————2

/2(n–2)

sections 6.7, 6.8, 7.7 (note: the approach used here to present the material in these sections is...

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