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5.14bb: How many renters must be included, if the confidence level is lowered to 90%? Hint: See L-11 pdf and lecture, page 6

5.14bb: How many renters must be included, if the confidence level is lowered to 90%? Hint: See L-11 pdf and lecture, page 6

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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QUESTION 3 5.14bb: How many renters must be included, if the confidence level is lowered to 90%? Hint: See L-11 pdf and lecture, page 6
Transcribed Image Text:QUESTION 3 5.14bb: How many renters must be included, if the confidence level is lowered to 90%? Hint: See L-11 pdf and lecture, page 6
EXAMPLE 5.14 Refer to Example 5.6. Using a preset value of a = .01, is there sufficient evidence in the data to support the research hypothesis? Solution The null and alternative hypotheses are Họ: p= 190 H u# 190 From the sample data, with s replacing o, the computed value of the test statistic is y- H. 178.2 – 190 aNn 45.3/N100 -2.60 The level of significance for this test statistic is computed using the formula given in Example 5.13, Case 3. p-value = 2P(z = |computed z) = 2P(z -2,60|) = 2P(z = 2.60) = 2(1 - .9953) = .0094 Because the p-value is very small, there is very little evidence to support H9. More precisely, p-value = .0094 s .01 = a, and, hence, we reject Hp. Thus, there is suf- ficient evidence (p-value = .0094) to support the research hypothesis and conclude that the mean cholesterol level differs from 190. Note that this is exactly the same conclusion reached using the traditional approach.
Transcribed Image Text:EXAMPLE 5.14 Refer to Example 5.6. Using a preset value of a = .01, is there sufficient evidence in the data to support the research hypothesis? Solution The null and alternative hypotheses are Họ: p= 190 H u# 190 From the sample data, with s replacing o, the computed value of the test statistic is y- H. 178.2 – 190 aNn 45.3/N100 -2.60 The level of significance for this test statistic is computed using the formula given in Example 5.13, Case 3. p-value = 2P(z = |computed z) = 2P(z -2,60|) = 2P(z = 2.60) = 2(1 - .9953) = .0094 Because the p-value is very small, there is very little evidence to support H9. More precisely, p-value = .0094 s .01 = a, and, hence, we reject Hp. Thus, there is suf- ficient evidence (p-value = .0094) to support the research hypothesis and conclude that the mean cholesterol level differs from 190. Note that this is exactly the same conclusion reached using the traditional approach.
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