Assume that sample data,based on two independent samples of size 25,give us 1 = 23,and s 2 = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom) for  2 -  1 ? A) (-2.40,22.40) B) (-4.57,24.57) C) (-4.96,24.96) D) (5.79,14.21) class=answers-bank-image d-inline rel=preload = 505, 1 = 23,and s 2 = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom) for  2 -  1 ? A) (-2.40,22.40) B) (-4.57,24.57) C) (-4.96,24.96) D) (5.79,14.21) class=answers-bank-image d-inline rel=preload = 515,s 1 = 23,and s 2 = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom) for  2 -  1 ?A) (-2.40,22.40)B) (-4.57,24.57)C) (-4.96,24.96)D) (5.79,14.21)

Question

Assume that sample data,based on two independent samples of size 25,give us  1  = 23,and s 2  = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom) for   2  -   1 ? A) (-2.40,22.40)  B) (-4.57,24.57)  C) (-4.96,24.96)  D) (5.79,14.21)  class=answers-bank-image d-inline rel=preload > = 505,  1  = 23,and s 2  = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom) for   2  -   1 ? A) (-2.40,22.40)  B) (-4.57,24.57)  C) (-4.96,24.96)  D) (5.79,14.21)  class=answers-bank-image d-inline rel=preload > = 515,s 1  = 23,and s 2  = 28.What is a 95% confidence interval (use the conservative value for the degrees of freedom) for   2  -   1 ?A) (-2.40,22.40)B) (-4.57,24.57)C) (-4.96,24.96)D) (5.79,14.21)

Jonathan Robles · Accepted Answer

Final Answer : C, Explanation :  We first need to calculate the sample variances for both samples:s1^2 = ((23-505)^2 + (28-505)^2)/24 = 421s2^2 = ((23-515)^2 + (28-515)^2)/24 = 521Then, we calculate the F-statistic:F = s2^2/s1^2 = 1.237Using 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 < 1.205Finally, we solve for the difference in means:515/28 - 505/25 = 2.321Multiplying by the confidence interval for F:0.837 * 2.321 < F1-F2 < 1.205 * 2.321Simplifying:-1.94 < F1-F2 < 2.79Therefore, 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 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 ₂ - ₁?

A) (-2.40,22.40)
B) (-4.57,24.57)
C) (-4.96,24.96)
D) (5.79,14.21)

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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 2 - 1?A) (-2.40,22.40) B) (-4.57,24.57) C) (-4.96,24.96) D) (5.79,14.21)

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Learning Objectives

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Related questions

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 ₂ - ₁?

A) (-2.40,22.40)
B) (-4.57,24.57)
C) (-4.96,24.96)
D) (5.79,14.21)