Students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Assume that a simple random sample has been selected. Use a 

0.100.10 

significance level to test the claim that these times are from a population with a mean equal to 

6060 

seconds. Does it appear that students are reasonably good at estimating one​ minute?

6969

8181

3939

6565

4141

2121

6060

6363
 
6565

4949

6262

6969

9797

9292

6767

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:

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  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 (\(s\)):

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

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

   \[
   s \approx 19.91
   \]

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

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

4. **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.

5. **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.

Learn how to perform a two-tailed t-test for the mean of a sample with a significance level of 0.10. This example involves testing if students' estimated times for one minute differ from the true mean of 60 seconds. The page covers hypothesis testing steps, including calculating the sample mean, standard deviation, and test statistic.

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