The 95 confidence interval is round to two decimal places a

The 95% confidence interval is (_____,_____)

round to two decimal places as needed.

Consider the following data from two independent samples with equal population variances. Construct a 95% confidence interval to estimate the difference in population means. Assume the population variances are equal and that the populations are normally distributed. X_1 vector = 67.8 X_2 vector = 75.1 s_1 = 12.9 s_2 = 8.3 n_1 = 10 n_2 = 14

Solution

Calculating the means of each group,              
              
X1 =    67.8          
X2 =    75.1          
              
Calculating the standard deviations of each group,              
              
s1 =    12.9          
s2 =    8.3          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    10          
n2 = sample size of group 2 =    14          
Thus, df = min(n1-1, n2-1) =    9
          
Also, sD =    4.643459302          
              
For the   0.9   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.05          
t(alpha/2) =    2.262157163          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -17.80423472          
upper bound = [X1 - X2] + t(alpha/2) * sD =    3.204234719          
              
Thus, the confidence interval is              
              
(   -17.80423472   ,   3.204234719   ) [ANSWER]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!