The past records of a supermarket show that its customers spend an average of $9

Question

The past records of a supermarket show that its customers spend an average of $95 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 encouraged to spend more money at the store. To check whether this is true the manager of the store took a sample of 14 customers who visited the store. The following data give the money (in dollars spent by these customers at this supermarket during their visits. 107.26 109.16 123.51 124.01 99.07 69.75 85.39 86.59 100.1 99.14 101.00 94.63 89.19 120.39 Assume that the money spent by a customers at this supermarket has a normal distribution. Using a 2.5% significance level, can you conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than $95? Round the sample standard deviation and sample mean to three decimal places. We conclude that the mean amount of money spent by all customers at this supermarket after the campaign was started is $95.

Explanation / Answer

From the given data we have

Below are the null and alternate hypothesis
H0: mu <= 95
H1 > 95

test statistics, t = (100.656 - 95)/(14.958/sqrt(14)) = 1.4148

p-value = 0.0903

As p-value is greater than significance level of 0.025, we fail to reject the null hypothesis.
This means money spent is not greater than 95

xbar 100.656 s 14.958

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