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WileyPLUS C Secure | https://edugen.vvileyplus.com/ed WileyPLUSa HyWileyPLUSI Hep I Contact Us I Log Out WileyPLUS an, Introductory statistics, 9e INTRO STATS. (MQM 100) Home Read, Study & Practice Gradebook ORION Downloadable eTextbook Assionment Open Assignment ASSIGNMENT RESOURCES Chapter 09, Section 9.3, Problem 048 The past records of a supermarket show that its customers spend an average of 90 per visit at this store Recently the management of the store initiate a promotional campaign according to which each customer receives points based on the total money spent at the store, and these points can be used to buy products at the store. The management expects that as a result of this campaign, the customers should be encouraged to spend more money at the store. To check whather this is true, tha manager ofthe store took a sample of 14 customers who visited the store. The follawing data give the money (in dallars) spent by these customers at this supermarket during their visits. 107.65 109.85 124,85 125.35 97.57 67.20 85.51 87.01 100.9 58.64 101.63 96.70 89.19 117.78 Assume that the money spent by all customers at this supermarket has a normal distrb on sin a 10% significance eve can you conclude that the mean amount o money spent by all customers at this supermarket after he campaign was started is more than $907 (Hint: First calculate the sample rnaan and the sample standard deviation for these data. Then make the test of hypothesis about .) Round the sample standerd deviation end sample mean to three decimal places We conclude that the mean amount of money spent by all customers at this supermarket after the campeign was started Question Attempts: 0 of 2 used SAVE FOR LATER SUBMIT ANSWER apter 09, Sect Verzion 4.2433 4 46 PM O Type here to search

Explanation / Answer

Given that,
population mean(u)=90
sample mean, x =100.7164
standard deviation, s =15.9751
number (n)=14
null, Ho: =90
alternate, H1: >90
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.65
since our test is right-tailed
reject Ho, if to > 2.65
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =100.7164-90/(15.9751/sqrt(14))
to =2.51
| to | =2.51
critical value
the value of |t | with n-1 = 13 d.f is 2.65
we got |to| =2.51 & | t | =2.65
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 2.51 ) = 0.01305
hence value of p0.01 < 0.01305,here we do not reject Ho
ANSWERS
---------------
null, Ho: =90
alternate, H1: >90
test statistic: 2.51
critical value: 2.65
decision: do not reject Ho
p-value: 0.01305
the mean amount of money spent by all customers at this super market is equal to $90

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