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The following MINITAB output (first shown in Exercise 16 in Section 6.1) presents the results of a hypothesis test for a population mean u. One-Sample Z: X Test of mu = 73.5 vs not = 73.5 The assumed standard deviation = 2.3634 Variable N Mean StDev SE Mean 95% CI х 145 73.2461 2.3634 0.1963 (72.8614, 73.6308) –1.29 0.196 a. Can Họ be rejected at the 5% level? How can you tell? b. Someone asks you whether the null hypothesis H,: H = 73 versus H1: H# 73 can be rejected at the 5% level. Can you answer without doing any calculations? How?

The following MINITAB output (first shown in Exercise 16 in Section 6.1) presents the results of a hypothesis test for a population mean u. One-Sample Z: X Test of mu = 73.5 vs not = 73.5 The assumed standard deviation = 2.3634 Variable N Mean StDev SE Mean 95% CI х 145 73.2461 2.3634 0.1963 (72.8614, 73.6308) –1.29 0.196 a. Can Họ be rejected at the 5% level? How can you tell? b. Someone asks you whether the null hypothesis H,: H = 73 versus H1: H# 73 can be rejected at the 5% level. Can you answer without doing any calculations? How?

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
Problem 1P
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The following MINITAB output (first shown in Exercise 16 in Section 6.1) presents the results of a hypothesis test for a population mean u. One-Sample Z: X Test of mu = 73.5 vs not = 73.5 The assumed standard deviation = 2.3634 Variable N Mean StDev SE Mean 95% CI х 145 73.2461 2.3634 0.1963 (72.8614, 73.6308) –1.29 0.196 a. Can Họ be rejected at the 5% level? How can you tell? b. Someone asks you whether the null hypothesis H,: H = 73 versus H1: H# 73 can be rejected at the 5% level. Can you answer without doing any calculations? How?
Transcribed Image Text:The following MINITAB output (first shown in Exercise 16 in Section 6.1) presents the results of a hypothesis test for a population mean u. One-Sample Z: X Test of mu = 73.5 vs not = 73.5 The assumed standard deviation = 2.3634 Variable N Mean StDev SE Mean 95% CI х 145 73.2461 2.3634 0.1963 (72.8614, 73.6308) –1.29 0.196 a. Can Họ be rejected at the 5% level? How can you tell? b. Someone asks you whether the null hypothesis H,: H = 73 versus H1: H# 73 can be rejected at the 5% level. Can you answer without doing any calculations? How?
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