A researcher wants to compare the variances of the heights (in inches) of 4 year college basketball players with those of players in junior colleges. A sample of 30 players from each type of school is selected, and the variances of the heights for each type are: 4 year colleges – 2.43 inches, junior colleges – 3.15 inches. Assuming the heights of all basketball players is normally distributed, at α = 0.10, is there a significant difference between the variances of the heights in the two types of schools

To determine if there is a significant difference between the variances of the heights in the two types of schools, we can perform a F-test, also known as the F-ratio test for comparing two variances.

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The variances of the heights in the two types of schools are equal.
- Alternative hypothesis (HA): The variances of the heights in the two types of schools are not equal.

Step 2: Determine the significance level (α):
The significance level given in the question is α = 0.10.

Step 3: Compute the F-statistic:
The formula for the F-statistic is:

F = s1^2 / s2^2

where s1^2 and s2^2 are the sample variances for the 4-year colleges and junior colleges, respectively.

In this case, s1^2 = 2.43 inches and s2^2 = 3.15 inches.

Plugging these values into the formula, we get:

F = 2.43 / 3.15

Step 4: Calculate the critical value:
To determine the critical value for the F-statistic, we need the degrees of freedom (df1 and df2) for the two samples. In this case, both samples have 30 players each, so both df1 and df2 are 30 - 1 = 29.

Using a significance level of α = 0.10 and the degrees of freedom (29 and 29), we can find the critical value from an F-distribution table or a statistical calculator. The critical value for α = 0.10 and df1 = 29 and df2 = 29 is approximately 1.6909.

Step 5: Compare the calculated F-statistic with the critical value:
If the calculated F-statistic is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference between the variances. If the calculated F-statistic is less than or equal to the critical value, we fail to reject the null hypothesis.

In this case, the calculated F-statistic is:

F = 2.43 / 3.15 ≈ 0.7714

Since 0.7714 < 1.6909, we fail to reject the null hypothesis.

Step 6: Interpret the results:
Based on the analysis, there is not enough evidence to conclude that there is a significant difference between the variances of the heights in four-year colleges and junior colleges.

To determine if there is a significant difference between the variances of the heights in the two types of schools, we can perform a hypothesis test. In this case, we will use the F-test.

The null hypothesis (H₀) is that the variances of the heights in the two types of schools are equal. The alternative hypothesis (H₁) is that the variances are not equal.

To perform the F-test, we calculate the test statistic, which is the ratio of the sample variances:

F = S₁² / S₂²

Where S₁² is the sample variance of the heights in 4-year colleges (2.43 inches), and S₂² is the sample variance of the heights in junior colleges (3.15 inches).

The degrees of freedom for the numerator (4-year colleges) is n₁ - 1 = 30 - 1 = 29, and the degrees of freedom for the denominator (junior colleges) is n₂ - 1 = 30 - 1 = 29.

Next, we look up the critical value of F for a significance level of α = 0.10 and the degrees of freedom (29, 29).

Finally, we compare the calculated F-value to the critical value. If the calculated F-value is greater than the critical value, we reject the null hypothesis and conclude that there is a significant difference between the variances. Otherwise, we fail to reject the null hypothesis.

You can use statistical software or a statistical table to find the critical value and calculate the F-value.

A researcher wants to compare the variances of the heights (in inches) of four-

years college basketball players with those of players in junior colleges. A sample of
30 players from each type of school is selected, and the variances of the heights for
each type are 2.43 and 3.15, respectively. At α = 0.10, is there a significant difference
between the variances of the heights in the two types of schools? Find the 90%
confidence interval for the difference of the variances.