Question

Is there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4687 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2121. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use

α = 0.10.

State the null and alternative hypotheses.

H₀: σ₁² = σ₂²

H_a: σ₁² ≠ σ₂²

H₀: σ₁² ≤ σ₂²

H_a: σ₁² > σ₂²

    

H₀: σ₁² ≠ σ₂²

H_a: σ₁² = σ₂²

H₀: σ₁² > σ₂²

H_a: σ₁² ≤ σ₂²

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the p-value. (Round your answer to four decimal places.)

p-value =

State your conclusion.

Reject H₀. We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers.Do not reject H₀. We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers.     Reject H₀. We can conclude that there is a difference in the variability of golf scores for male and female professional golfers.Do not reject H₀. We can conclude that there is a difference in the variability of golf scores for male and female professional golfers.

Answer 1

H₀: $\sigma1^2 = \sigma2^2$

H₁: $\sigma1^2 \neq \sigma2^2$

The test statistic F = $\dpi{100} \frac{s_{1}^2}{s_{2}^2}$

                           $\dpi{100} = \frac{(2.4687)^2}{(2.2121)^2}$

                           = 1.25

P-value = 2 * 0.2872

           = 0.5744

Since the P-value is greater than the significance level(0.5744 > 0.10), so we should not reject the null hypothesis.

Do not reject H₀. We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers.