13. If a random sample of 20 homes in area A has a mean selling price of $128,000 and a standarddeviation of $4500, and a random sample of 20 homes in area B has a mean selling price of$125,000 and a standard deviation of $2500 can you conclude that there is a significantdifference between the variability in the selling price of homes in these two areas at the 0.05level?a) Solve using the P-value approach.b) Solve using the critical value approach
Question
- If a random sample of 20 homes in area A has a mean selling price of 4500, and a random sample of 20 homes in area B has a mean selling price of2500 can you conclude that there is a significantdifference between the variability in the selling price of homes in these two areas at the 0.05level?a) Solve using the P-value approach.b) Solve using the critical value approach
Solution
a) P-value approach:
Step 1: State the null hypothesis and the alternative hypothesis. Null hypothesis (H0): σ1^2 = σ2^2 (The variances are equal) Alternative hypothesis (H1): σ1^2 ≠ σ2^2 (The variances are not equal)
Step 2: Calculate the test statistic. The test statistic for the equality of variances is the ratio of the larger sample variance to the smaller sample variance. In this case, the test statistic F = s1^2 / s2^2 = 4500^2 / 2500^2 = 3.24.
Step 3: Determine the P-value. The P-value is the probability that a random variable is greater than the observed test statistic. Using the F-distribution table with degrees of freedom df1 = n1 - 1 = 19 and df2 = n2 - 1 = 19, we find that the P-value is less than 0.05.
Step 4: Make a decision. Since the P-value is less than the significance level of 0.05, we reject the null hypothesis. Therefore, we can conclude that there is a significant difference between the variability in the selling price of homes in these two areas.
b) Critical value approach:
Step 1: State the null hypothesis and the alternative hypothesis. Null hypothesis (H0): σ1^2 = σ2^2 (The variances are equal) Alternative hypothesis (H1): σ1^2 ≠ σ2^2 (The variances are not equal)
Step 2: Determine the critical value. The critical value for a two-tailed test at the 0.05 level with degrees of freedom df1 = n1 - 1 = 19 and df2 = n2 - 1 = 19 is approximately 2.54 (from the F-distribution table).
Step 3: Calculate the test statistic. The test statistic F = s1^2 / s2^2 = 4500^2 / 2500^2 = 3.24.
Step 4: Make a decision. Since the test statistic is greater than the critical value, we reject the null hypothesis. Therefore, we can conclude that there is a significant difference between the variability in the selling price of homes in these two areas.
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