Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Assume that both populations are normally distributed.

a. Test whether u1≠ u2 at the alpha=0.05 level of signifigance for the given sample data. (u= population mean, sorry couldnt insert the symbol). Determine p value. Should the null hypothesis be rejected?
b. Construct a 95% confidence interval about μ1−μ2. at the alphα=0.05 level of significance for the given sample data.

Population 1 Population 2
n 18 18
x 12.7 14.6
s 3.2 3.8


Sagot :

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]

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.