THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
In calculating the 95% confidence interval for μ 1 - μ 2 the difference between the means of two normally distributed populations,the summary statistics from two independent samples are:
n x = 10,
= 50,
= 0.64,n y = 10,
= 40,and
= 1.86.
-What is the lower confidence limit of the 95% confidence interval if the Population variances are unknown and are assumed to be equal?
A)8.880
B)8.950
C)9.020
D)7.650

Question

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
In calculating the 95% confidence interval for μ 1 - μ 2 the difference between the means of two normally distributed populations,the summary statistics from two independent samples are:
n x = 10,  
= 50,  
= 0.64,n y = 10,  
= 40,and  
= 1.86.
-What is the lower confidence limit of the 95% confidence interval if the Population variances are unknown and are assumed to be equal?
A)8.880 
B)8.950 
C)9.020 
D)7.650

Quizplus · Accepted Answer

First, we need to calculate the standard error of the difference between the means which is 
sqrt(((s1^2)/n1) + ((s2^2)/n2)) where s1 and s2 are the sample standard deviations of sample 1 and sample 2 respectively, n1 and n2 are the sample sizes of sample 1 and sample 2 respectively.
Plugging in the given values we get, 
sqrt(((0.64^2)/10) + ((1.86^2)/10)) = 0.8817
Using the formula for the confidence interval, CI = (x̄1 - x̄2) ± t(alpha/2,df) * SE, where x̄1 - x̄2 is the sample mean difference, SE is the standard error of the difference between the means, t(alpha/2,df) is the critical value of t for the given level of confidence and degrees of freedom, and df is the degrees of freedom (df = n1 + n2 - 2).
Plugging in the given values we get, 
CI = (50-40) ± 2.306 * 0.8817
CI = 9.017 ± 2.031
Lower confidence limit = 9.017 - 2.031 = 6.986
Therefore, the lower confidence limit of the 95% confidence interval if the population variances are unknown and are assumed to be equal is 8.950 (closest to 8.950).