Assume that Friday morning​ taxi-cab rides have times with a standard deviation of sigmaequals10.2 minutes. A cab driver records times of rides during a Friday afternoon time period and obtains these​ statistics: nequals11​, x overbarequals18.4 ​minutes, sequals13.9 minutes. Use a 0.05 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 Friday afternoon times have greater variation than Friday morning times, we conduct a hypothesis test for the variance using an \( F \)-test. Here are the steps:

### Step 1: Define the Hypotheses
Let:
- \(\sigma^2\) represent the variance of Friday afternoon times,
- \(\sigma_0^2 = 10.2^2 = 104.04\) represent the variance of Friday morning times.

The hypotheses are:
- **Null Hypothesis** \( H_0 \): \( \sigma^2 = \sigma_0^2 \) (Friday afternoon times have the same or lower variance than Friday morning times),
- **Alternative Hypothesis** \( H_1 \): \( \sigma^2 > \sigma_0^2 \) (Friday afternoon times have greater variance than Friday morning times).

### Step 2: Set the Significance Level
The significance level \( \alpha = 0.05 \).

### Step 3: Compute the Test Statistic
We use the formula for the test statistic \( F \):
\[
F = \frac{s^2}{\sigma_0^2}
\]
where:
- \( s = 13.9 \) is the sample standard deviation of Friday afternoon times,
- \( \sigma_0 = 10.2 \).

Calculating \( s^2 \) and then the \( F \)-statistic:
\[
s^2 = 13.9^2 = 193.21
\]
\[
F = \frac{193.21}{104.04} \approx 1.857
\]

### Step 4: Determine the Critical Value
Since we are conducting a one-tailed test, we find the critical \( F \)-value for \( \alpha = 0.05 \), with degrees of freedom:
- \( df_{\text{numerator}} = n - 1 = 11 - 1 = 10 \),
- \( df_{\text{denominator}} = \infty \) (or large-sample approximation for Friday morning times).

Using an \( F \)-distribution table or calculator, we find that the critical value for \( F \) at \( df = 10 \) and a one-tailed \( \alpha = 0.05 \) is approximately \( F_{\text{critical}} = 2.978 \).

### Step 5: Make a Decision
- If \( F \geq F_{\text{critical}} \), reject \( H_0 \).
- If \( F < F_{\text{critical}} \), fail to reject \( H_0 \).

Since \( 1.857 < 2.978 \), we **fail to reject \( H_0 \)**.

### Conclusion
At a 0.05 significance level, there is insufficient evidence to support the claim that Friday afternoon taxi-cab ride times have greater variation than Friday morning times.

Would you like further details on any steps or clarification? 

Here are five related questions to extend your understanding:
1. How would the result change if the sample size were larger?
2. What would happen if the significance level was changed to 0.01?
3. How do we interpret the \( F \)-distribution in terms of variance?
4. What assumptions are critical for the validity of this \( F \)-test?
5. How would this test differ if we wanted to test for a smaller variance instead?

**Tip:** The \( F \)-test for variance is sensitive to normality. Small deviations from a normal distribution can affect the test outcome.

This solution explains how to test the claim that Friday afternoon taxi-cab ride times have greater variation than Friday morning times using an F-test for variance. Given statistics for a sample size of 11, sample mean of 18.4 minutes, and sample standard deviation of 13.9 minutes, the solution walks through the hypothesis testing steps with a significance level of 0.05. Ideal for students learning statistical inference and hypothesis testing at the college level.

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