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 3.11 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. If the test statistic is 2.2, what is the p-value associated with this hypothesis test?(Round your answer to three decimal places.)

Answer 1

Answer:

At 0.05 significance level, the p-value is 0.014

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 3 minutes

Sample mean, $\bar{x}$ = 3.11 minutes

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, σ = 0.5 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 3\text{ minutes}\\H_A: \mu > 3\text{ minutes}$

We use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

$z_{stat} = 2.2$

Now, we calculate the p-value from the normal standard z-table.

P-value = 0.014

At 0.05 significance level, the p-value is 0.014