Answered: lest the P-value an d decide wrhethei,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Complete parts (a) and (b) below.
Click here to view Page 1 of the table of areas under the standard normal curve.
Click here to view Page 2 of the table of areas under the standard normal curve.
a. The test statistic in a right-tailed test is z= 1.71. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject
the null hypothesis in favor of the alternative hypothesis.
The P-value is
(Round to three decimal places as needed.)
This P-value
V sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is
V the significance level.
b. The test statistic in a right-tailed test is z= - 0.82. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to
reject the null hypothesis in favor of the alternative hypothesis.
The P-value is
(Round to three decimal places as needed.)
This P-value
sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is
V the significance level.

Expert Solution

Step 1

The required P-value can be obtained by using the standard normal table. For this case, find the area under the standard normal curve to the left of the z score of 1.71. The required area is 0.044. Hence, the required P-value is 0.044. It is known that if the p-value is greater than the level of significance, the null hypothesis does not get rejected. As the p-value (0.044) is less than the level of significance (0.05), the null hypothesis is rejected.

(a)

The p-value is 0.044

This p-value provides sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is less than the significance.

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