Question

Given two independent random samples with the following results:

n1=6

x‾1=192

s1=12

n2=9

x‾2=162

s2=31

Use this data to find the 90 % confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 3: Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 3 of 3: Construct the 90% confidence interval. Round your answers to the nearest whole number.

Answer 1

1)

Given CI level is 0.9, hence α = 1 - 0.9 = 0.1                  
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.771

2)

Pooled Variance
sp = sqrt((((n1 - 1)*s1^2 + (n2 - 1)*s2^2)/(n1 + n2 - 2))*(1/n1 + 1/n2))
sp = sqrt((((6 - 1)*12^2 + (9 - 1)*31^2)/(6 + 9 - 2))*(1/6 + 1/9))
sp = 13.4037
                  
                  
Margin of Error                  
ME = tc * sp                  
ME = 1.771 * 13.4037                  
ME = 23.737953                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (192 - 162 - 1.771 * 13.4037 , 192 - 162 - 1.771 * 13.4037                  
CI = (6.262 , 53.738)                  
CI = (6 , 54)