Math Problem Statement

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.

Solution

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 FF-test. Here are the steps:

Step 1: Define the Hypotheses

Let:

  • σ2\sigma^2 represent the variance of Friday afternoon times,
  • σ02=10.22=104.04\sigma_0^2 = 10.2^2 = 104.04 represent the variance of Friday morning times.

The hypotheses are:

  • Null Hypothesis H0H_0: σ2=σ02\sigma^2 = \sigma_0^2 (Friday afternoon times have the same or lower variance than Friday morning times),
  • Alternative Hypothesis H1H_1: σ2>σ02\sigma^2 > \sigma_0^2 (Friday afternoon times have greater variance than Friday morning times).

Step 2: Set the Significance Level

The significance level α=0.05\alpha = 0.05.

Step 3: Compute the Test Statistic

We use the formula for the test statistic FF: F=s2σ02F = \frac{s^2}{\sigma_0^2} where:

  • s=13.9s = 13.9 is the sample standard deviation of Friday afternoon times,
  • σ0=10.2\sigma_0 = 10.2.

Calculating s2s^2 and then the FF-statistic: s2=13.92=193.21s^2 = 13.9^2 = 193.21 F=193.21104.041.857F = \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 FF-value for α=0.05\alpha = 0.05, with degrees of freedom:

  • dfnumerator=n1=111=10df_{\text{numerator}} = n - 1 = 11 - 1 = 10,
  • dfdenominator=df_{\text{denominator}} = \infty (or large-sample approximation for Friday morning times).

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

Step 5: Make a Decision

  • If FFcriticalF \geq F_{\text{critical}}, reject H0H_0.
  • If F<FcriticalF < F_{\text{critical}}, fail to reject H0H_0.

Since 1.857<2.9781.857 < 2.978, we fail to reject H0H_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 FF-distribution in terms of variance?
  4. What assumptions are critical for the validity of this FF-test?
  5. How would this test differ if we wanted to test for a smaller variance instead?

Tip: The FF-test for variance is sensitive to normality. Small deviations from a normal distribution can affect the test outcome.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Variance
F-distribution
Statistical Inference

Formulas

F = s^2 / sigma_0^2

Theorems

F-test for variance

Suitable Grade Level

College/Undergraduate