Question

9.48 The past records of a supermarket show that its customers spend an average of $95 per visit at this store. Recently the manage- ment 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 109.15 136.01 107.02 116.15 101.53 109.29 110.79 94.83 100.91 97.94 104.30 83.54 67.59 120.44 Assume that the money spent by all customers at this supermarket has a normal distribution. Using a 5% significance level, can you con- clude that the mean amount of money spent by all customers at this supermarket after the campaign was started is more than $95? (Hint: First calculate the sample mean and the sample standard deviation for these data using the formulas learned in Sections 3.1.1 and 3.2.2 of Chapter 3. Then make the test of hypothesis about u.)

Answer 1

Solution:

We are given that sample of 14 customers who visited the store and we have to test if the mean amount of money spent by all customers at this supermarket after the campaign was started is more than $95.

Level of significance = 5% = 0.05

Find sample mean and sample standard deviation for the given sample.

x	x^2
109.15	11913.72
136.01	18498.72
107.02	11453.28
116.15	13490.82
101.53	10308.34
109.29	11944.3
110.79	12274.42
94.83	8992.729
100.91	10182.83
97.94	9592.244
104.3	10878.49
83.54	6978.932
67.59	4568.408
120.44	14505.79

Thus mean is:

and sample standard deviation is:

Now use following steps for hypothesis testing:

State H0 and H1:

Vs

Step 2) Test statistic:

We use t test statistic:

Step 3) t critical value:

df = n - 1 =14 -1 = 13

One tail area = Level of significance = 5% = 0.05

t critical value = 1.771

Step 4) Rejection rule: Reject H0 , if t test statistic value > t critical value, otherwise we fail to reject H0.

Since t test statistic value = 2.130 > t critical value = 1.771 , we reject H0.

Step 5) Conclusion: There is sufficient evidence to support the claim that : the mean amount of money spent by all customers at this supermarket after the campaign was started is more than $95.