Question

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known at 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly different than 3 minutes. Use a = 0.05. The test statistic is a 1.96 b. 1.64 c. 2.00 d. 0.056 The p-value is a. 0.0456 b. 0.0482 c. 0.4772 d. 0.9759 e. 0.9772 At the 5% level, you a. fail to reject the null hypothesis b. reject the null hypothesis

Answer 1

Answer:

The test statistic is c. 2.00

The p-value is a. 0.0456

At the 5% level, you b. reject the null hypothesis

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly different than 3 minutes.

This means that the mean and the alternate hypothesis are:

Null: $H_{0} = 3$

Alternate: $H_{a} = 3$

The test-statistic is given by:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

Sample of 100 customers.

This means that $n = 100$

3 tested at the null hypothesis

This means that $\mu = 3$

The average length of time it took the customers in the sample to check out was 3.1 minutes.

This means that $X = 3.1$

The population standard deviation is known at 0.5 minutes.

This means that $\sigma = 0.5$

Value of the test-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}$

$z = 2$

The test statistic is z = 2.

The p-value is

Mean different than 3, so the pvalue is 2 multiplied by 1 subtracted by the pvalue of Z when z = 2.

z = 2 has a pvalue of 0.9772

2*(1 - 0.9772) = 2*0.0228 = 0.0456

At the 5% level

0.0456 < 0.05, which means that the null hypothesis is rejected.