The past records of a supermarket show that its customers spend an average of $4

Question

The past records of a supermarket show that its customers spend an average of $45 per visit at this store. Recently the management of the store initiated 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 to spend more money at the store. To check whether this is true, the manager of the store took a sample of customers who visited the store. The following data give the money (in dollars) spent by these customers at this supermarket during their visits. 88 69 141 28 106 45 32 51 78 54 110 83 The mean of this data is x = 73.8, and the standard deviation is s = 34.1 Assume that the money spent by all customers at this supermarket has a normal distribution Answer the following questions. (a) Construct a 90% confidence interval for the mean based on this sample. (b) Provide a valid statistical interpretation of this confidence interval. For parts c to g of this problem conduct a hypothesis test at a 5% significance level to determine if the data supports the claim that the mean amount of money spent by all customers at this supermarket is higher than $45 after the campaign by answering the following questions. (c) State the null and alternative hypothesis using proper notation and indicate if left, right, or 2-Tailed test. (d) Justify and state the proper distribution and test statistic for conducting this test. (E) Calculate the value for the test statistic indicated in proceeding part. Determine the correct critical value or p-value, depending on your choice of method. (G) State and justify the conclusion as a mathematical statement well as plain English a interpretation. A sketch is not required, but may be helpful in explaining your reasoning.

Explanation / Answer

Part-a

Degree of freedom=n-1=12-1=11

Crititcal t for 90% confidence tc =1.796

90% CI= xbar±tc*s/sqrt(n)

=73.8±1.796*34.1/sqrt(12)

=(56.12 91.48)

Part-b

We are 90% confident that this interval contains the true population mean

Part-c

Null hypothesis H0:µ=45

Alternative hypothesis Ha: µ>45

This is right tailed test

Part-d

The distribution is Student’s t-distribution with n-1 degree of freedom as we do not know the population standard deviation and sample size is small and population is assumed to be normal

Part-e

Test statistic t=(xbar-45)/(s/sqrt(n))

=(73.8-45)/(34.1/sqrt(12))

=2.926

Part-f

Degree of freedom =n-1=12-1=11

One tailed right p-value=tdist(2.926,11,1)= 0.0069

Part-g

As p-value,0.05 we reject the null hypothesis and conclude that mean amount of money spent by all customers is higher than $45.

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