Question

The population mean waiting time to check out of a supermarket has been 4 minutes. Recently, in an effort to reduce the waiting time, the supermarket has experimented with a system in which infrared cameras use body heat and in-store software to determine how many lanes should be opened. A sample of 100 customers was selected, and their mean waiting time to check out was 3.10 minutes, with a sample standard deviation of 2.5 minutes. At the 0.10 significance level is there evidence that the population mean waiting for time to check out is less than 4 minutes?
a. State the null and alternate hypotheses
b. Find the critical value
c. Compute the value of the test statistic
d. What is your decision regarding the null hypothesis? Interpret the result.

Answer 1

Answer:

We conclude that the population mean waiting for time to check out is less than 4 minutes.

Step-by-step explanation:

We are given that the population mean waiting time to check out of a supermarket has been 4 minutes.

A sample of 100 customers was selected, and their mean waiting time to check out was 3.10 minutes, with a sample standard deviation of 2.5 minutes.

Let $\mu$ = population mean waiting for time to check out.

So, Null Hypothesis, $H_0$ : $\mu$ $\geq$ 4 minutes     {means that the population mean waiting for time to check out is more than or equal to 4 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ < 4 minutes     {means that the population mean waiting for time to check out is less than 4 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

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

where, $\bar X$ = sample mean waiting time to check out = 3.10 minutes

            $\sigma$ = sample standard deviation = 2.5 minutes

            n = sample of customers = 100

So, the test statistics  =  $\frac{3.10-4}{\frac{2.5}{\sqrt{100} } }$  ~ $t_9_9$

                                     =  -3.6

The value of t test statistics is -3.6.

Now, at 0.10 significance level the t table gives critical value of -1.291 at 99 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -3.6 < -1.291, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the population mean waiting for time to check out is less than 4 minutes.