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 0.10
 
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?
70
83
39
66
41
21
60
62
 
65
47
66
69
95
91
67
	
Question content area bottom
Part 1
Perform the test assuming that the requirements are met. Identify the null and alternative hypotheses.
Upper H 0
​: 
mu
equals
60Upper H 1
​: 
mu
not equals
60
​(Type integers or decimals. Do not​ round.)
Part 2
Identify the test statistic.
enter your response here
​(Round to two decimal places as​ needed.)

To test the claim that the times are from a population with a mean equal to 60 seconds, we'll conduct a one-sample t-test. Here's the step-by-step solution:

### Part 1: Hypotheses

The null and alternative hypotheses are:

- **Null Hypothesis (\(H_0\))**: \(\mu = 60\) (The mean time is equal to 60 seconds.)
- **Alternative Hypothesis (\(H_1\))**: \(\mu \neq 60\) (The mean time is not equal to 60 seconds.)

### Part 2: Test Statistic

First, we need to calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (\(s\)).

Given the dataTo test the claim that the times are from a population with a mean equal to 60 seconds, we'll conduct a one-sample t-test. Here's the step-by-step solution:

### Part 1: Hypotheses

The null and alternative hypotheses are:

- **Null Hypothesis (\(H_0\))**: \(\mu = 60\) (The mean time is equal to 60 seconds.)
- **Alternative Hypothesis (\(H_1\))**: \(\mu \neq 60\) (The mean time is not equal to 60 seconds.)

### Part 2: Test Statistic

First, we need to calculate the sample mean (\(\bar{x}\)) and the sample standard deviation (\(s\)).

Given the 
\[ 70, 83, 39, 66, 41, 21, 60, 62, 65, 47, 66, 69, 95, 91, 67 \]

**Step 1: Calculate the sample mean (\(\bar{x}\))**

\[
\bar{x} = \frac{\sum x_i}{n}
\]
Where:
- \(n = 15\) (the number of observations)
- \(\sum x_i\) is the sum of all observations.

**Step 2: Calculate the sample standard deviation (\(s\))**

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

**Step 3: Calculate the test statistic (t)**

\[
t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}
\]
Where:
- \(\mu_0 = 60\) (the population mean under the null hypothesis)

I'll now calculate these values.### Results:

- **Sample Mean (\(\bar{x}\))**: 62.8 seconds
- **Sample Standard Deviation (\(s\))**: 19.68 seconds (rounded to two decimal places)
- **Test Statistic (t)**: 0.55 (rounded to two decimal places)

### Conclusion:

Using the calculated test statistic of \(t = 0.55\), you can now compare this with the critical value from the t-distribution table for a two-tailed test at the 0.10 significance level with \(n-1 = 14\) degrees of freedom, or directly find the p-value associated with this t-statistic.

Do you want to proceed with finding the p-value or critical value to finalize the hypothesis test? Let me know if you need further details!

---

Here are five related questions you might want to explore:

1. How do you determine the degrees of freedom in a t-test?
2. What is the p-value, and how do you interpret it?
3. How does the significance level affect the outcome of a hypothesis test?
4. What assumptions must be met for the t-test to be valid?
5. How would the test results change if the sample size were larger?

**Tip:** When conducting a hypothesis test, always ensure the assumptions (e.g., normality, independence) are reasonably met to validate your results.

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 0.10 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? 70, 83, 39, 66, 41, 21, 60, 62, 65, 47, 66, 69, 95, 91, 67.

Test the claim that students can accurately estimate the length of one minute using a one-sample t-test at a 0.10 significance level. The dataset includes 15 student estimates, and the solution involves calculating the sample mean, standard deviation, and t-test statistic to determine if the mean is equal to 60 seconds.

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