Chapter 10.3, Problem 88E

(a)

To determine

To explain why or why not can you conclude that the difference in shopping behavior is due to the effect of Friday the $13\text{th}$ on people’s behavior.

Answer to Problem 88E

No.

Explanation of Solution

As we know that the selected stores are a random sample while we took two Fridays in the same month.

However, it is possible that some special event occurs in either of these two weeks. For example, the $4\text{th}$ of July is the national holiday in the U.S. and thus it is possible that the more people take the reminder of the week off. This then implies that the higher number of shoppers on the $6\text{th}$ could be due to the national holiday of the $4\text{th}$ of July instead of due to a different behavior on the Friday the $13\text{th}$ .

(b)

To determine

To explain do these data provide convincing evidence at the $\alpha =0.05$ level that the number of shoppers at the grocery stores on these two days differs on average.

Answer to Problem 88E

There is no convincing evidence at the $\alpha =0.05$ level that the number of shoppers at the grocery stores on these two days differs on average.

Explanation of Solution

It is given that:

  $\begin{array}{l}\alpha =0.05\\ n=45\\ {\overline{x}}_D=-46.5\\ s_D=178.0\end{array}$

Given claim is that: Difference.

The claim is either null hypothesis or an alternative hypothesis.

  $\begin{array}{l}{H}_0:{\mu }_D=0\\ H_a:{\mu }_D\ne 0\end{array}$

The test statistics value will be:

  $\begin{array}{l}t=\frac{{\overline{x}}_D-{\mu }_D}{\frac{s_D}{\sqrt{n}}}\\ =\frac{-46.5-0}{\frac{178.0}{\sqrt{45}}}\\ =-1.752\end{array}$

And the degree of freedom will be:

  $\begin{array}{l}df=n-1\\ =45-1\\ =44\end{array}$

As we know that the P -value is the probability of obtaining the sample results or results more extreme when the null hypothesis is true.

  $0.05=2(0.025)<P<2(0.05)=0.10$

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we conclude that there is no convincing evidence at the $\alpha =0.05$ level that the number of shoppers at the grocery stores on these two days differs on average.

(c)

To determine

To explain what type of error could you have made- Type I error or Type II error, based on the conclusion in part (a).

Answer to Problem 88E

Type II error.

Explanation of Solution

We conclude in part (a) and part (b) that,

There is convincing evidence to support the claim that that the number of shoppers at the grocery stores on these two days differs on average.

A type I error occurs if we reject a null hypothesis when the null hypothesis is true. And the Type II error occurs if we fails to reject the null hypothesis when the null hypothesis is false.

Thus, in this case we fails to reject the null hypothesis then it is a Type II error.