Two samples are taken from different populations, one with a sample size of [tex]n_1 = 5[/tex] and the other with a sample size of [tex]n_2 = 11[/tex]. The mean of the first sample is [tex]\bar{X}_1 = 37.9[/tex] and the mean of the second sample is [tex]\bar{X}_2 = 406.3[/tex], with variances [tex]s_1^2 = 64.2[/tex] and [tex]s_2^2 = 135.1[/tex], respectively. Can we conclude that the variances of the two populations differ, using [tex]\alpha = 0.05[/tex]?

The null hypothesis (H0) states that the variances of the two populations are equal, while the alternative hypothesis (Ha) states that the variances are different.

The test statistic for the F-test is calculated as the ratio of the sample variances: F = s12 / s22.

For the given sample data, we have s12 = 64.2 and s22 = 135.1. Plugging these values into the formula, we get F ≈ 0.475.

To conduct the hypothesis test, we compare the calculated F-value to the critical F-value. The critical value is determined based on the significance level (α) and the degrees of freedom for the two samples.

In this case, α = 0.05 and the degrees of freedom for the two samples are (n1 - 1) = 4 and (n2 - 1) = 10, respectively.

Using an F-table or a calculator, we can find the critical F-value with α = 0.05 and degrees of freedom (4, 10) to be approximately 4.26.

Since the calculated F-value (0.475) is less than the critical F-value (4.26), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.

Note that the conclusion may change if a different significance level is chosen.

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jhasagar3210 jhasagar3210 · Answer 2 · 2025-02-23T20:50:49-05:00

We do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.

To determine whether the variances of the two populations differ, we can perform a hypothesis test using the F-test.

The null hypothesis (H0) states that the variances of the two populations are equal, while the alternative hypothesis (Ha) states that the variances are different.

The test statistic for the F-test is calculated as the ratio of the sample variances: F = s12 / s22.

For the given sample data, we have s12 = 64.2 and s22 = 135.1. Plugging these values into the formula, we get F ≈ 0.475.

To conduct the hypothesis test, we compare the calculated F-value to the critical F-value. The critical value is determined based on the significance level (α) and the degrees of freedom for the two samples.

In this case, O = 0.05 and the degrees of freedom for the two samples are (n1 - 1) = 4 and (n2 - 1) = 10, respectively.

Using an F-table or a calculator, we can find the critical F-value with o = 0.05 and degrees of freedom (4, 10) to be approximately 4.26.

Since the calculated F-value (0.475) is less than the critical F-value (4.26), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the variances of the two populations differ at the 0.05 significance level.

Note that the conclusion may change if a different significance level is chosen.

Learn more about variances from below link

brainly.com/question/9304306

#SPJ11