Math Problem Statement

Question content area top 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.

Solution

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 (σ\sigma) = 10.4 minutes
  • Friday Afternoon:
    • Sample size (nn) = 13
    • Sample mean (x\overline{x}) = 19.4 minutes
    • Sample standard deviation (ss) = 12.7 minutes
  • Significance level (α\alpha) = 0.01

Hypotheses:

  • Null Hypothesis (H0H_0): σafternoon2σmorning2\sigma^2_{\text{afternoon}} \leq \sigma^2_{\text{morning}}
  • Alternative Hypothesis (HaH_a): σafternoon2>σmorning2\sigma^2_{\text{afternoon}} > \sigma^2_{\text{morning}}

Test Statistic:

The test statistic for the F-test is calculated as:

F=safternoon2σmorning2F = \frac{s^2_{\text{afternoon}}}{\sigma^2_{\text{morning}}}

Calculating:

  1. safternoon2=(12.7)2=161.29s^2_{\text{afternoon}} = (12.7)^2 = 161.29
  2. σmorning2=(10.4)2=108.16\sigma^2_{\text{morning}} = (10.4)^2 = 108.16

Substituting into the F statistic formula:

F=161.29108.161.49F = \frac{161.29}{108.16} \approx 1.49

Critical Value:

Using an F-distribution table, we find the critical value for α=0.01\alpha = 0.01 with degrees of freedom df1=n1=12df_1 = n - 1 = 12 (afternoon sample) and df2df_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 F0.01,12,F_{0.01, 12, \infty} is approximately 2.67.

Conclusion:

  • Compare FF to the critical value: 1.49<2.671.49 < 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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
Variance Comparison
F-Test

Formulas

F = s^2 / σ^2
s^2 = (Sample Standard Deviation)^2
σ^2 = (Population Standard Deviation)^2

Theorems

F-distribution for testing equality of variances

Suitable Grade Level

Undergraduate Statistics