Answer:
Fail to reject the null hypothesis
[tex]CI = (-4.278, 0.478)[/tex]
Step-by-step explanation:
Given
[tex]n_1=n_2 = 18[/tex]
[tex]\bar x_1 = 12.7[/tex] [tex]\bar x_2 = 14.6[/tex]
[tex]\sigma_1 = 3.2[/tex] [tex]\sigma_2 = 3.8[/tex]
[tex]\alpha = 0.05[/tex]
Solving (a): Test the hypothesis
We have:
[tex]H_o : \mu_1 - \mu_2 = 0[/tex]
[tex]H_a : u1 - u2 \ne 0[/tex]
Calculate the pooled standard deviation
[tex]s_p = \sqrt\frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2}{n_1+n_2-2}}[/tex]
[tex]s_p = \sqrt\frac{(18-1)*3.2^2 + (18-1)*3.8^2}{18+18-2}}[/tex]
[tex]s_p = \sqrt\frac{419.56}{34}}[/tex]
[tex]s_p = \sqrt{12.34}[/tex]
[tex]s_p = 3.51[/tex]
Calculate test statistic
[tex]t = \frac{x_1 - x_2}{s_p*\sqrt{1/n_1 + 1/n_2}}[/tex]
[tex]t = \frac{12.7 - 14.6}{3.51 *\sqrt{1/18 + 1/18}}[/tex]
[tex]t = \frac{-1.9}{3.51 *\sqrt{1/9}}[/tex]
[tex]t = \frac{-1.9}{3.51 *1/3}[/tex]
[tex]t = \frac{-1.9}{1.17}[/tex]
[tex]t = -1.62[/tex]
From the t table, the p value is:
[tex]p = 0.114472[/tex]
[tex]p > \alpha[/tex]
i.e.
[tex]0.114472 > 0.05[/tex]
So, the conclusion is that: we fail to reject the null hypothesis.
Solving (b): Construct 95% degree freedom
[tex]\alpha = 0.05[/tex]
Calculate the degree of freedom
[tex]df = n_1 + n_2 - 2[/tex]
[tex]df = 18+18 - 2[/tex]
[tex]df = 34[/tex]
From the student t table, the t value is:
[tex]t = 2.032244[/tex]
The confidence interval is calculated as:
[tex]CI = (x_1 - x_2) \± s_p * t * \sqrt{1/n_1 + 1/n_2}[/tex]
[tex]CI = (12.7 - 14.6) \± 3.51 * 2.032244 * \sqrt{1/18 + 1/18}[/tex]
[tex]CI = (12.7 - 14.6) \± 3.51 * 2.032244 * \sqrt{1/9}[/tex]
[tex]CI = (12.7 - 14.6) \± 3.51 * 2.032244 * 1/3[/tex]
[tex]CI = -1.90 \± 2.378[/tex]
Split
[tex]CI = (-1.90 - 2.378, -1.90 + 2.378)[/tex]
[tex]CI = (-4.278, 0.478)[/tex]
