A coffee shop chain wants to know if the average time customers spend in their stores during peak hours is less than 45 minutes. They sampled 20 customers and recorded their time spent in the store. We need to perform a hypothesis test to determine if there is sufficient evidence to support the claim that the average time is less than 45 minutes, given that the population standard deviation $\sigma$ is unknown. We will use a hypothetical value for the sample standard deviation.

1. Problem Description

A coffee shop chain wants to know if the average time customers spend in their stores during peak hours is less than 45 minutes. They sampled 20 customers and recorded their time spent in the store. We need to perform a hypothesis test to determine if there is sufficient evidence to support the claim that the average time is less than 45 minutes, given that the population standard deviation $\sigma$ is unknown. We will use a hypothetical value for the sample standard deviation.

2. Solution Steps

Step 1: State the null and alternative hypotheses.

The null hypothesis ($H_0$) is that the average time ($\mu$) is greater than or equal to 45 minutes.

The alternative hypothesis ($H_1$) is that the average time ($\mu$) is less than 45 minutes.

$H_0: \mu \ge 45$

$H_1: \mu < 45$

Step 2: Select the level of significance ($\alpha$).

Let's choose a significance level of $\alpha = 0.05$. This means that we are willing to accept a 5% chance of rejecting the null hypothesis when it is actually true.

Step 3: Choose the appropriate test statistic and justify your selection.

Since the population standard deviation ($\sigma$) is unknown and the sample size ($n=20$) is less than 30, we should use the t-test. The t-test is appropriate when the population standard deviation is unknown and the sample size is small.

The test statistic is:

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

where:

$\bar{x}$ is the sample mean

$\mu_0$ is the hypothesized population mean (45 in this case)

$s$ is the sample standard deviation

$n$ is the sample size

Step 4: Calculate the value of the test statistic.

Let's assume a hypothetical sample mean $\bar{x} = 42$ minutes and a sample standard deviation $s = 8$ minutes. Then, the test statistic is:

$t = \frac{42 - 45}{8 / \sqrt{20}} = \frac{-3}{8 / \sqrt{20}} = \frac{-3}{8 / 4.472} = \frac{-3}{1.789} = -1.677$

Thus, the calculated test statistic t = -1.677

Step 5: Formulate the decision rule based on the critical value and the p-value.

Since this is a left-tailed test with $\alpha = 0.05$ and degrees of freedom $df = n - 1 = 20 - 1 = 19$, we need to find the critical t-value ($t_{critical}$) from the t-distribution table.

Looking up the t-table for $\alpha = 0.05$ (one-tailed) and $df = 19$, we find $t_{critical} = -1.729$.

The decision rule is: Reject $H_0$ if $t < t_{critical}$.

Alternatively, we can use the p-value approach. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true. Using a t-distribution calculator or software, the p-value for $t = -1.677$ with $df = 19$ for a left-tailed test is approximately $0.054$.

The decision rule is: Reject $H_0$ if p-value < $\alpha$.

Step 6: Decide based on your decision rule.

Based on the critical value approach: Since our test statistic $t = -1.677$ is not less than the critical value $t_{critical} = -1.729$, we fail to reject the null hypothesis.

Based on the p-value approach: Since the p-value $0.054$ is greater than $\alpha = 0.05$, we fail to reject the null hypothesis.

Step 7: Interpret the decision.

We fail to reject the null hypothesis. There is not enough evidence to support the claim that the average time customers spend at the coffee shop during peak hours is less than 45 minutes at a significance level of 0.05, given our hypothetical data.

3. Final Answer

There is not enough evidence to support the claim that the average time customers spend at the coffee shop during peak hours is less than 45 minutes at a significance level of 0.

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