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 0.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one​ minute?          69, 80, 38, 64, 41, 21, 59, 62, 65, 49, 64, 71, 94, 89, 65                           Identify the test statistic. Find p-value                                                                                                                         

To determine if the students' times are reasonably close to 60 seconds on average, we'll perform a **one-sample t-test**. This test will help us assess if the sample mean significantly deviates from 60 seconds at a 0.01 significance level.

Here’s the step-by-step solution:

### Step 1: Set up Hypotheses
1. **Null Hypothesis (H₀)**: The mean time is equal to 60 seconds.
   \[
   H_0: \mu = 60
   \]
2. **Alternative Hypothesis (H₁)**: The mean time is not equal to 60 seconds.
   \[
   H_1: \mu \neq 60
   \]

### Step 2: Sample Mean and Standard Deviation
Calculate the sample mean \(\bar{x}\) and sample standard deviation \(s\) based on the given times.

- **Sample Data**: \(69, 80, 38, 64, 41, 21, 59, 62, 65, 49, 64, 71, 94, 89, 65\)

1. **Sample Size (n)**: \(n = 15\)
2. **Sample Mean (\(\bar{x}\))**: Compute \(\bar{x}\).
3. **Sample Standard Deviation (s)**: Compute \(s\).

### Step 3: Test Statistic
The test statistic \(t\) for a one-sample t-test is calculated by:
\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\]
where:
- \(\mu = 60\) (population mean under the null hypothesis),
- \(\bar{x}\) is the sample mean,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

### Step 4: Find the p-value
Using the calculated \(t\)-value and degrees of freedom (\(df = n - 1 = 14\)), find the p-value for a two-tailed test.

### Step 5: Compare p-value to Significance Level
If the p-value is less than 0.01, we reject the null hypothesis, indicating that the sample mean is significantly different from 60 seconds.

I'll calculate these values for you.### Results

1. **Sample Mean (\(\bar{x}\))**: \(62.07\) seconds
2. **Sample Standard Deviation (s)**: \(19.10\) seconds
3. **Test Statistic (t)**: \(0.419\)
4. **p-value**: \(0.682\)

### Conclusion
Since the p-value \(0.682\) is much greater than the significance level of \(0.01\), we **fail to reject the null hypothesis**. This suggests that there is no significant evidence to conclude that the average estimated time by students is different from 60 seconds.

### Interpretation
It appears that students are reasonably accurate in estimating one minute, as their mean estimation time does not significantly deviate from 60 seconds.

---

Do you want more details or have any questions? 

### Related Questions
1. What would the outcome be if the significance level were 0.05?
2. How would the result change if the sample size were larger?
3. Why is a t-test appropriate here instead of a z-test?
4. What assumptions are made in this one-sample t-test?
5. How sensitive is the t-test to outliers in the sample data?

### Tip
When dealing with small sample sizes, the t-distribution provides a more reliable estimate than the normal distribution, especially if the population standard deviation is unknown.

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 0.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute? 69, 80, 38, 64, 41, 21, 59, 62, 65, 49, 64, 71, 94, 89, 65 Identify the test statistic. Find p-value.

Explore a one-sample t-test to determine if students can accurately estimate one minute without a clock. Given a sample of times, learn how to calculate the test statistic, p-value, and interpret results at a 0.01 significance level. Sample data includes 69, 80, 38, 64, and more.

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