Question

A marketing researcher wants to estimate the mean amount spent per year​ ($)on a web site by membership member shoppers. Suppose a random sample of 100membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $56and a standard deviation of$55.Complete parts​(a)and​ (b)below.

a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from $53?(Usea 0.01 level of​significance.)

State the null and alternative hypotheses.

Identify the critical values (round to two decimals)

Determine the test statistic: tSTAT​= (round to two decimals)

(Reject/ Do not Reject) H0. There is (sufficient/insufficient)evidence that the population mean spent by membership member customers is different from $53.

b. determine the p- value (round to 3 decimal places)

Interpret the meaning of the​ p-value.Select the correct answer below.

A.The​ p-valueis the probability of obtaining a sample mean that is equal to or more extreme than

​$3 away from $53 if the null hypothesis is true.

B.The​ p-valueis the probability of obtaining a sample mean that is equal to or more extreme than

​$3 above $53 if the null hypothesis is false.

C.The​ p-valueis the probability of not rejecting the null hypothesis when it is false.

D.The​ p-valueis the probability of obtaining a sample mean that is equal to or more extreme than

​$3 below $53 if the null hypothesis is false.

Answer 1

given that

sample size=n=100

sample mean =m=56

sample SD=S=55

a)

we have to test that population mean is different from 53 or not hence

since test is two tailed so

here we will use t statistics with df=n-1=100-1=99

critical values for alpha =0.01

P(t< left tailed critical value)=0.005 from t table P(t<-2.63) =0.005

Hence left tailed critical value =-2.63

right tailed critical value is given by

P(t> right tailed critical value) =0.005 from t table P(t>2.63) =0.005

hence right tailed critical value=2.63

Hence critical values are -2.63,2.63

now test statistics is given by

since t stat lies between 2.63 and -2.63 hence failed to reject H0 hence

Do not RejectH0. There is Insufficientevidence that the population mean spent by membership member customers is different from $53.

b)

Test is two tailed so

P-Value =2*P(t>0.54)=2*0.295 =0.590

A.The​ p-valueis the probability of obtaining a sample mean that is equal to or more extreme than

​$3 away from $53 if the null hypothesis is true.