A store manager is interested in testing the claim that the proportion of online sales that are over $500 is different than23%. To test this claim, the store manager collects the following data on a sample of 720 online sales and 180 of them areover $500. The following is the data from this study:Sample size = 500 online sales• The alternative hypothesis is H. :p# 0.23• The test statistic is calculated as zn = 1.28.Find and interpret the p- value for this hypothesis test.Use the curve below to help visualize your answer. Select the appropriate test by dragging the blue point to a right-, left- ortwo-tailed diagram, then set the test statistic on the x-axis to find the p-value.Based on the p-value, choose the correct interpretation.Move the blue dot to choose the appropriate testp-value= 0.2006-1test statistic = 1.28desmosAAA Move the blue dot to choose the appropriate testp-value= 0.2006test statistic =1.282powered bydesmosSelect the correct answer below:O The probability of observing a value of zo = 1.28 or less if the null hypothesis is true is 20.06%.O The probability of observing a value of zo = ±1.28 or more extreme if the null hypothesis is true is 20.06%.O The probability of observing a value of z0 = 1.28 or less if the null hypothesis is true is 79.94%.O The probability of observing a value of zo = ±1.28 or more extreme if the null hypothesis is true is 79.94%.O FEEDBACKMORE INSTRUCTIONSUBMIT

Name: A store manager is interested in testing the claim that the proportio
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Description: A store manager is interested in testing the claim that the proportion of online sales that are over $500 is different than23%. To test this claim, the store manager collects the following data on a sample of 720 online sales and 180 of them areover

The test is two tailed, and test statistic is 1.28. The p-value for two tailed test is, p-value=Pz<-1.28 or z>1.28=2×Pz<-1.28 The probability of z less than –1.28 can be obtained using the excel formula “=NORM.S.DIST(–1.28,TRUE)”. The probability value is 01003. The required p-value is, p-value=2×Pz<-1.28=2×0.1003=0.2006 Thus, the p-value is 0.2006. The appropriate graph is,The probability of observing a value z0=±1.28 or more extreme if the null hypothesis is true is 20.06%

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Find and interpret the p- value for this hypothesis test. Use the curve below to help visualize your answer. Select the appropriate test by dragging the blue point to a right-, left- or two-tailed diagram, then set the test statistic on the x-axis to find the p-value.

Find and interpret the p- value for this hypothesis test. Use the curve below to help visualize your answer. Select the appropriate test by dragging the blue point to a right-, left- or two-tailed diagram, then set the test statistic on the x-axis to find the p-value.

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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