Question

The owners of a shopping mall had studied the shopping habits of their customers. Historical data showed that the mean time spent by shoppers in the mall was 60 minutes. The owners improved the mall and wished the improvement would cause shoppers to spend more time in the mall. A random sample of 50 shoppers is selected, the average time spent by these shoppers in the mall is 62 minutes with a standard deviation of 5 minutes.

(a) Use p-value approach, is there evidence to suggest that the improvement has caused shoppers to spend more time in the mall at 5% level of significance?

(b) Explain why the sampling distribution of sample mean follow the normal distribution in this case.

Answer 1

[We first answer part (b) and then (a), as justification given in (b) is required to perform the test as done in (a) here.]

(b)

We do not know population variance.

In such case, in general, we have to perform t-test for testing of hypothesis for population mean.

But here n = 50 and we know that for n > 30, t-distribution can be approximated to normal distribution.

Hence, for n = 50 (>30) this sampling distribution of sample mean follows the normal distribution in this case.

(a)

We have to test for null hypothesis

against the alternative hypothesis

The test statistic is given by

Here,   

Alternative hypothesis is of greater than type. So it is a right tail test.

Corresponding P value = 0.002339066    [Using R code "1-pnorm(2.8284)"]

Level of significance

We reject null hypothesis if P value < Level of significance

Since , we reject our null hypothesis.

Based on given observation, we can conclude that is there is sufficient evidence to suggest that the improvement has caused shoppers to spend more time in the mall.