A pizza delivery chain advertises that it will deliver your pizza in 2525 minutes from when the order is placed. Being a skeptic, you decide to test and see if the mean delivery time is actually more than 2525 minutes. For the simple random sample of 1212 customers who record the amount of time it takes for each of their pizzas to be delivered, the mean is 27.627.6 minutes with a standard deviation of 4.24.2 minutes. Assume that the population distribution is approximately normal. Perform a hypothesis test using a 0.050.05 level of significance.

Step 3 of 3 :

Draw a conclusion and interpret the decision.

Let's go through the hypothesis testing process step by step to draw a conclusion and interpret the decision.

### **Given Data:**
- Sample size (\(n\)) = 12
- Sample mean (\(\bar{x}\)) = 27.6 minutes
- Sample standard deviation (\(s\)) = 4.2 minutes
- Hypothesized population mean (\(\mu_0\)) = 25 minutes
- Significance level (\(\alpha\)) = 0.05

### **Step 1: State the Hypotheses**

We are conducting a one-tailed hypothesis test to determine if the mean delivery time is greater than 25 minutes.

- **Null Hypothesis** (\(H_0\)): \(\mu = 25\) (the mean delivery time is 25 minutes)
- **Alternative Hypothesis** (\(H_a\)): \(\mu > 25\) (the mean delivery time is greater than 25 minutes)

### **Step 2: Calculate the Test Statistic**

We will use the **t-test** because the sample size is small (\(n = 12\)) and the population standard deviation is unknown.

The formula for the t-test statistic is:

\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]

Where:
- \(\bar{x} = 27.6\) (sample mean)
- \(\mu_0 = 25\) (hypothesized mean)
- \(s = 4.2\) (sample standard deviation)
- \(n = 12\) (sample size)

Substituting the values:

\[
t = \frac{27.6 - 25}{\frac{4.2}{\sqrt{12}}} = \frac{2.6}{1.213} \approx 2.14
\]

### **Step 3: Determine the Critical Value**

Since the test is one-tailed with a significance level of \(0.05\), we need to find the critical value of \(t\) for \(n - 1 = 12 - 1 = 11\) degrees of freedom from the t-distribution table.

For \(df = 11\) and \(\alpha = 0.05\), the critical value is approximately **1.796**.

### **Step 4: Make a Decision**

Now, compare the calculated test statistic \(t = 2.14\) with the critical value \(t_{0.05, 11} = 1.796\):

- Since \(2.14 > 1.796\), we **reject the null hypothesis**.

### **Step 5: Conclusion and Interpretation**

The test statistic is greater than the critical value, so we reject the null hypothesis. This means that, at the 0.05 significance level, there is sufficient evidence to conclude that the mean delivery time is **significantly greater than 25 minutes**.

### Relative Questions:
1. What if the standard deviation had been larger? How would that affect the test result?
2. What if the sample size were larger? How would the test statistic change?
3. What happens if we change the significance level from 0.05 to 0.01? Would the decision remain the same?
4. How does the shape of the t-distribution change as the sample size increases?
5. If we conducted a two-tailed test instead, what would the null and alternative hypotheses be?

**Tip:** When performing hypothesis testing, always consider whether the test should be one-tailed or two-tailed based on the research question.

A pizza delivery chain advertises that it will deliver your pizza in 25 minutes from when the order is placed. Being a skeptic, you decide to test and see if the mean delivery time is actually more than 25 minutes. For the simple random sample of 12 customers who record the amount of time it takes for each of their pizzas to be delivered, the mean is 27.6 minutes with a standard deviation of 4.2 minutes. Assume that the population distribution is approximately normal. Perform a hypothesis test using a 0.05 level of significance.

Learn how to perform a hypothesis test to determine if the mean pizza delivery time exceeds 25 minutes. This example uses a t-test with a sample of 12 customers, a mean of 27.6 minutes, and a standard deviation of 4.2 minutes. Suitable for undergraduate statistics students.

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