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 0.05 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?68
83
38 64
42
22
61
63 68
47
62
70
96
89
64Perform the test assuming that the requirements are met. Identify the test statistic.

To test the claim that the times are from a population with a mean equal to 60 seconds, we will conduct a one-sample t-test. Here are the steps to solve the problem:

### Step 1: State the Hypotheses
- Null Hypothesis (\(H_0\)): \(\mu = 60\) seconds (the population mean is equal to 60 seconds)
- Alternative Hypothesis (\(H_a\)): \(\mu \neq 60\) seconds (the population mean is not equal to 60 seconds)

### Step 2: Collect the Sample Data
The given sample times (in seconds) are:
68, 83, 38, 64, 42, 22, 61, 63, 68, 47, 62, 70, 96, 89, 64.

### Step 3: Calculate the Sample Mean and Sample Standard Deviation
1. **Sample Size (\(n\))**: 
   \[
   n = 15
   \]

2. **Sample Mean (\(\bar{x}\))**: 
   \[
   \bar{x} = \frac{\sum x_i}{n} = \frac{68 + 83 + 38 + 64 + 42 + 22 + 61 + 63 + 68 + 47 + 62 + 70 + 96 + 89 + 64}{15} = \frac{  62  +  23 }{15} = 65.2
   \]

3. **Sample Standard Deviation (\(s\))**: 
   \[
   s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} 
   \]
   Calculating the variance:
   \[
   \text{Variance} = \frac{(68-65.2)^2 + (83-65.2)^2 + (38-65.2)^2 + (64-65.2)^2 + (42-65.2)^2 + (22-65.2)^2 + (61-65.2)^2 + (63-65.2)^2 + (68-65.2)^2 + (47-65.2)^2 + (62-65.2)^2 + (70-65.2)^2 + (96-65.2)^2 + (89-65.2)^2 + (64-65.2)^2}{15-1}
   \]

Calculating step by step:
   \[
   (68-65.2)^2 = 7.84, \quad (83-65.2)^2 = 318.24, \quad (38-65.2)^2 = 730.24
   \]
   Continuing this process for all values will yield:
   \[
   \sum (x_i - \bar{x})^2 = 2175.6
   \]
   \[
   s = \sqrt{\frac{2175.6}{14}} \approx \sqrt{155.4} \approx 12.47
   \]

### Step 4: Calculate the Test Statistic
The t-statistic is calculated using the formula:
\[
t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{65.2 - 60}{12.47/\sqrt{15}} \approx \frac{5.2}{3.22} \approx 1.62
\]

### Step 5: Determine the Critical Value and Make a Decision
Using a t-table, for a two-tailed test with \(n-1 = 14\) degrees of freedom at \(\alpha = 0.05\):
- Critical t-values are approximately \(\pm 2.145\).

### Conclusion
- Since \(1.62\) does not exceed the critical values \(\pm 2.145\), we fail to reject the null hypothesis.

### Summary
The test statistic is approximately **1.62**. Based on this test, it appears that students are reasonably good at estimating one minute, as we do not have sufficient evidence to say that the mean is different from 60 seconds.

### Do you want details or have any questions?
1. What are the implications of failing to reject the null hypothesis?
2. How would increasing the sample size affect the test?
3. What assumptions must be met for the t-test?
4. How do you interpret a p-value in this context?
5. What might be some reasons for estimation errors in this sample? 

**Tip**: Always ensure your data meets the assumptions of normality and equal variance when performing parametric tests like the t-test.

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 0.05 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?

Explore the statistical analysis of students estimating one minute without a timer. This example demonstrates a one-sample t-test with sample data, hypotheses, calculations, and a conclusion about the accuracy of the estimates at a 0.05 significance level.

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