To test the claim that the students' times are from a population with a mean equal to 60 seconds, we will perform a hypothesis test for the population mean. Here are the steps involved:

1. Identify the null and alternative hypotheses:

   • Null hypothesis ($H_0$): $\mu = 60$ (The mean time is 60 seconds)
   • Alternative hypothesis ($H_1$): $\mu \neq 60$ (The mean time is not 60 seconds)

2. Calculate the sample mean ($\bar{x}$) and sample standard deviation ($s$):

   • Sample dataTo test the claim that the students' times are from a population with a mean equal to 60 seconds, we will perform a hypothesis test for the population mean. Here are the steps involved:

3. Identify the null and alternative hypotheses:

   • Null hypothesis ($H_0$): $\mu = 60$ (The mean time is 60 seconds)
   • Alternative hypothesis ($H_1$): $\mu \neq 60$ (The mean time is not 60 seconds)

4. Calculate the sample mean ($\bar{x}$) and sample standard deviation ($s$):

   • Sample 69, 81, 39, 65, 41, 21, 60, 63, 65, 49, 62, 69, 97, 92, 67
   • Sample size ($n$): 15

   $\bar{x} = \frac{69 + 81 + 39 + 65 + 41 + 21 + 60 + 63 + 65 + 49 + 62 + 69 + 97 + 92 + 67}{15}$ $\bar{x} = \frac{940}{15} \approx 62.67$

   Now calculate the sample standard deviation ($s$):

   $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

   $s = \sqrt{\frac{(69-62.67)^2 + (81-62.67)^2 + (39-62.67)^2 + \cdots + (67-62.67)^2}{14}}$

   $s \approx 19.91$

5. Calculate the test statistic (t): $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ where $\mu_0 = 60$

   $t = \frac{62.67 - 60}{19.91 / \sqrt{15}}$ $t = \frac{2.67}{5.14} \approx 0.52$

6. Determine the P-value: Using the t-distribution table or a calculator, find the P-value for $t = 0.52$ with $df = n - 1 = 14$.

   For $t = 0.52$ and $df = 14$, the P-value for a two-tailed test is approximately 0.61.

7. State the final conclusion:

   • Significance level ($\alpha$): 0.10
   • Since the P-value (0.61) is greater than the significance level (0.10), we fail to reject the null hypothesis.

   The results do not provide sufficient evidence to conclude that the mean time is different from 60 seconds. It appears that students are reasonably good at estimating one minute.

Summary:

• Null Hypothesis ($H_0$): $\mu = 60$
• Alternative Hypothesis ($H_1$): $\mu \neq 60$
• Test Statistic: $t = 0.52$
• P-value: 0.61
• Conclusion: Fail to reject the null hypothesis. The sample data suggests that the mean time is not significantly different from 60 seconds.

Would you like more details or have any questions?

Here are five questions you might want to ask:

1. How do you calculate the test statistic for a t-test step-by-step?
2. What are the assumptions for conducting a t-test?
3. Can you explain the concept of P-value in hypothesis testing?
4. How do you interpret the results of a hypothesis test?
5. What is the difference between one-tailed and two-tailed tests?

Tip: Always verify the assumptions of normality and random sampling before performing a t-test to ensure the validity of the test results.