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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I just need to know why this is wrong below: What is the test statistic W? W=5 (incorrect) and  What is the p-value of this test? (p-value < 0.001-- incorrect)   Use the Wilcoxon signed rank test to test the hypothesis that the median number of pages in the statistics books in the library from which the sample was taken is 400. A sample of 12 statistics books have the following numbers of pages pages 127 217 486 132 397 297 396 327 292 256 358 272 What is the sum of the negative ranks (W-)? 75 What is the sum of the positive ranks (W+)? 5What type of test is this?     two tailedWhat is the test statistic W? 5 These are the critical values for a 1-tailed Wilcoxon Signed Rank test for n=12 Alpha Level 0.001 0.005 0.01 0.025 0.05 0.1 0.2 Critical Value 75 70 68 64 60 56 50 What is the p-value for this test?         p-value < 0.001
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