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Part 1
Assume that Friday morning​ taxi-cab rides have times with a standard deviation of sigmaequals10.4 minutes. A cab driver records times of rides during a Friday afternoon time period and obtains these​ statistics: nequals13​, x overbarequals19.4 ​minutes, sequals12.7 minutes. Use a 0.01 significance level to test the claim that these Friday afternoon times have greater variation than the Friday morning times. Assume that the sample is a simple random sample selected from a normally distributed population.

To test the claim that the variation in Friday afternoon taxi-cab ride times is greater than that of Friday morning times, we can perform an F-test for equality of variances. 

### Given:
- **Friday Morning**: 
  - Population standard deviation (\( \sigma \)) = 10.4 minutes
- **Friday Afternoon**: 
  - Sample size (\( n \)) = 13
  - Sample mean (\( \overline{x} \)) = 19.4 minutes
  - Sample standard deviation (\( s \)) = 12.7 minutes
- **Significance level (\( \alpha \))** = 0.01

### Hypotheses:
- Null Hypothesis (\( H_0 \)): \( \sigma^2_{\text{afternoon}} \leq \sigma^2_{\text{morning}} \)
- Alternative Hypothesis (\( H_a \)): \( \sigma^2_{\text{afternoon}} > \sigma^2_{\text{morning}} \)

### Test Statistic:
The test statistic for the F-test is calculated as:

\[
F = \frac{s^2_{\text{afternoon}}}{\sigma^2_{\text{morning}}}
\]

Calculating:

1. \( s^2_{\text{afternoon}} = (12.7)^2 = 161.29 \)
2. \( \sigma^2_{\text{morning}} = (10.4)^2 = 108.16 \)

Substituting into the F statistic formula:

\[
F = \frac{161.29}{108.16} \approx 1.49
\]

### Critical Value:
Using an F-distribution table, we find the critical value for \( \alpha = 0.01 \) with degrees of freedom \( df_1 = n - 1 = 12 \) (afternoon sample) and \( df_2 \) which is based on the population standard deviation (in this case, it's assumed to be infinite).

From the F-distribution table:
- Critical value for \( F_{0.01, 12, \infty} \) is approximately 2.67.

### Conclusion:
- Compare \( F \) to the critical value: \( 1.49 < 2.67 \).
- Since the test statistic does not exceed the critical value, we fail to reject the null hypothesis.

### Final Result:
At the 0.01 significance level, there is not enough evidence to conclude that the variation in Friday afternoon taxi-cab ride times is greater than that of Friday morning times.

Do you want more details on any part of this process? Here are some related questions you might consider:
1. What assumptions are made when performing an F-test?
2. How would the conclusion change if the significance level were higher?
3. What are the implications of failing to reject the null hypothesis in practical terms?
4. How can sample size affect the results of variance tests?
5. What are alternative tests for comparing variances? 

Tip: Always check the underlying assumptions of the statistical test being used, as they can influence the validity of the results.

Assume that Friday morning​ taxi-cab rides have times with a standard deviation of sigma=10.4 minutes. A cab driver records times of rides during a Friday afternoon time period and obtains these​ statistics: n=13​, x̄=19.4 ​minutes, s=12.7 minutes. Use a 0.01 significance level to test the claim that these Friday afternoon times have greater variation than the Friday morning times. Assume that the sample is a simple random sample selected from a normally distributed population.

Conduct a hypothesis test using an F-test to determine if Friday afternoon taxi-cab ride times have greater variation than Friday morning rides. Given a sample size of 13, a sample mean of 19.4 minutes, and a sample standard deviation of 12.7 minutes, learn how to apply the F-test at a 0.01 significance level.

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