The Basic Practice of Statistics
The Basic Practice of Statistics
7th Edition
ISBN: 9781464142536
Author: David S. Moore, William I. Notz, Michael A. Fligner
Publisher: W. H. Freeman
Question
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Chapter 17, Problem 17.44E

a.

To determine

To obtain: The 90% confidence interval for the mean blood pressure of all executives.

a.

Expert Solution
Check Mark

Answer to Problem 17.44E

The 90% confidence interval for the mean blood pressure of all executives is 123.162 to 128.978.

Explanation of Solution

Given info:

A medical director found mean blood pressure x¯=126.07 for an SRS of 72 executives with the standard deviation of the blood pressure of all executives as σ=15 .

Calculation:

The 90% confidence interval for the mean blood pressure of all executives is calculated by the formula,

Interval=x¯±zσn (1)

For the confidence interval 90, the value of z score is 1.645.

Substitute 126.07 for x¯ , 15 for σ , 72 for n and 1.645 for z in equation (1).

x¯±zσn=126.07±(1.645)(1572)=126.07±(1.645)(1.767)=126.07±2.907=123.162to128.978

Thus, the 90% confidence interval for the mean blood pressure of all executives is 123.162 to 128.978.

b.

To determine

To obtain: The z test for H0:μ=128 against the two sided alternative.

b.

Expert Solution
Check Mark

Answer to Problem 17.44E

The z test for H0:μ=128 against the two sided alternative is 1.0918 .

Explanation of Solution

Given Info:

The hypothesized value μ0=128 falls inside the confidence level.

Calculation:

The formula to calculate the z score is,

z=x¯μσn (2)

Substitute 126.07 for x¯ , 15 for σ , 72 for n and 128 for μ in equation (2).

z=126.071281572=1.931.76=1.0918

Thus the value of z score is 1.0918 .

Since the value of z score comes out to be 1.0918 , and by standard mean cumulative proportion table, the value lies between z=1.036 and z=1.282 , that means the P-value lies between P=0.30 and P=0.20 .

The level of significance is given as 10% that is α=0.1 and since the P value is greater than the level of significance thus it is concluded that the test is not significant on the 10% level.

c.

To determine

To obtain: The z test for H0:μ=129 against the two sided alternative.

c.

Expert Solution
Check Mark

Answer to Problem 17.44E

The z test for H0:μ=129 against the two sided alternative is 1.66 .

Explanation of Solution

Given Info:

The hypothesized value μ0=129 falls outside the confidence level.

Calculation:

The formula to calculate the z score is,

z=x¯μσn (2)

Substitute 126.07 for x¯ , 15 for σ , 72 for n and 129 for μ in equation (2) of part (b).

z=126.071291572=2.931.76=1.66

Thus the value of z score is 1.66 .

Since the value of z score comes out to be 1.66 , thus by standard mean cumulative proportion table, the value lies between z=1.645 and z=1.96 that means the P value lies between P=0.05 and P=0.10 .

The level of significance is given as 10% that is α=0.1 and since the P value is less than the level of significance thus it is concluded that the test is significant on the 10% level.

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