Question

The null and alternate hypotheses are:

H0 : μ1 = μ2
H1 : μ1 ≠ μ2
A random sample of 8 observations from one population revealed a sample mean of 23 and a sample standard deviation of 3.8. A random sample of 9 observations from another population revealed a sample mean of 27 and a sample standard deviation of 3.8. The population standard deviations are unknown but assumed to be equal. At the 0.05 significance level, is there a difference between the population means?

A. State the decision rule
B. Compute the pooled estimate of the population variance. (Round your answer to 3 decimal places.)
C. Compute the test statistic

Answer 1

a. The decision rule for a two-sample t-test with equal variances at the 0.05 significance level involves comparing the calculated t-value to the critical t-value.

b. The pooled estimate of the population variance is 14.4

c. The calculated test statistic is approximately -2.171.

How to calculate test statistics

State the decision rule:

The decision rule for a two-sample t-test with equal variances at the 0.05 significance level is as follows:

If the absolute value of the calculated test statistic is greater than the critical value (t_critical) with (n1 + n2 - 2) degrees of freedom, we reject the null hypothesis.

If the absolute value of the calculated test statistic is less than or equal to the critical value, we fail to reject the null hypothesis.

The pooled estimate of the population variance formula is:

`s_p^2 = ((n_1 - 1) * s_1^2 + (n_2 - 1) * s_2^2) / (n_1 + n_2 - 2)`

Given:

Sample 1: n_1 = 8, sample mean x_1 = 23, sample standard deviation s_1 = 3.8

Sample 2: n_2 = 9, sample mean x_2 = 27, sample standard deviation s_2 = 3.8

Let's compute the pooled estimate of the population variance:

`s_p^2 = ((8 - 1) * 3.8^2 + (9 - 1) * 3.8^2) / (8 + 9 - 2)`

`s_p^2` = (7 * 14.44 + 8 * 14.44) / 15

`s_p^2` = (100.08 + 115.52) / 15

`s_p^2` = 215.6 / 15

`s_p^2` ≈ 14.373

Thus, The pooled estimate of the population variance is 14.4

Test Statistic:

The formula for the test statistic in a two-sample t-test assuming equal variances is:

`t = (x_1 - x_2) / \sqrt(s_p^2 * (1/n_1 + 1/n_2))`

Given:

x_1 = 23, x_2 = 27

`s_p^2` ≈ 14.373 (pooled estimate of the population variance)

n_1 = 8, n_2 = 9

Let's compute the test statistic:

t = (23 - 27) /
`\sqrt`(14.373 * (1/8 + 1/9))

t = -4 /
`\sqrt`(14.373 * (0.125 + 0.111))

t = -4 /
`\sqrt`(14.373 * 0.236)

t = -4 /
`\sqrt`(3.396)

t ≈ -2.171 (rounded to three decimal places)

Therefore, the calculated test statistic is approximately -2.171.