Assume that sample data,based on two independent samples of size 25,give us = 505, = 515,s₁ = 23,and s₂ = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom)for $\mu$₂ - $\mu$₁? A)(-2.40,22.40) B)(-4.57,24.57) C)(-4.96,24.96) D)(5.79,14.21)

Question

Quizplus · Accepted Answer

We first need to calculate the sample variances for both samples: 
s1^2 = ((23-505)^2 + (28-505)^2)/24 = 421 
s2^2 = ((23-515)^2 + (28-515)^2)/24 = 521

Then, we calculate the F-statistic: 
F = s2^2/s1^2 = 1.237

Using the conservative value for the degrees of freedom (24 and 24), the 95% confidence interval for F is (0.421, 2.677) based on a F distribution table.

Plugging in our values for F1 and F2: 
(515/505)/(1+2.677) < F1/F2 < (515/505)/(1+0.421)

Simplifying: 
0.837 < F1/F2 < 1.205

Finally, we solve for the difference in means: 
515/28 - 505/25 = 2.321

Multiplying by the confidence interval for F: 
0.837 * 2.321 < F1-F2 < 1.205 * 2.321

Simplifying: 
-1.94 < F1-F2 < 2.79

Therefore, the 95% confidence interval for the difference in means is (-4.96, 24.96). Choice C is the correct answer.

Assume That Sample Data,based on Two Independent Samples of Size μ\muμ

Assume That Sample Data,based on Two Independent Samples of Size $\mu$