Question

The owner of a shoe store wanted to determine whether the average customer bought more than $100 worth of shoes. She randomly selected 10 receipts and identified the total spent by each customer. The totals (rounded to the nearest dollar) are given below.
Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether the mean is greater than $100 and then draw a conclusion in the context of the problem. Use α=0.05.
125 99 219 65 109 89 79 119 95 135
Select the correct answer below:
A) Reject the null hypothesis. There is sufficient evidence to conclude that the mean is greater than $100.
B) Reject the null hypothesis. There is insufficient evidence to conclude that the mean is greater than $100.
C) Fail to reject the null hypothesis. There is sufficient evidence to conclude that the mean is greater than $100.
D) Fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean is greater than $100.

Answer 1

Answer:

D) Fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean is greater than $100.

Step-by-step explanation:

We are given that the owner of a shoe store randomly selected 10 receipts and identified the total spent by each customer. The totals (rounded to the nearest dollar) are given below;

X: 125, 99, 219, 65, 109, 89, 79, 119, 95, 135.

Let $\mu$ = average customer bought worth of shoes.

So, Null Hypothesis, $H_0$ : $\mu \leq$ $100      {means that the mean is smaller than or equal to $100}

Alternate Hypothesis, $H_A$ : $\mu$ > $100      {means that the mean is greater than $100}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

                            T.S.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = $113.4

             s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $42.78

             n = sample of receipts = 10

So, the test statistics =  $\frac{113.4-100}{\frac{42.78}{\sqrt{10} } }$  ~  $t_9$

                                    =  0.991

The value of t-test statistics is 0.991.

Now, at a 0.05 level of significance, the t table gives a critical value of 1.833 at 9 degrees of freedom for the right-tailed test.

Since the value of our test statistics is less than the critical value of t as 0.991 < 1.833, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean is smaller than or equal to $100.