Question

Assuming the wing lengths of a subspecies of dark-eyed juncos to be normally distributed, consider samples of sizes 14 from two different subspecies. It is given that the two sample means are 82.1 and 84.9 and the two standard deviations are 1.501 and 1.698.

10a- Then what would be standard error of the point estimate of the difference of the mean wing lengths?

10b-What would be the degree of freedom of the t distribution in problem no. 10 above?

10c-What would be the t critical value for a 95% confidence interval in problem 10 above?

10d-What would be the Margin of Error for the 95% Confidence Interval above?

10e- The required 95% confidence interval in this problem is

Answer 1

Part a)

Standard Error = SP / √( (1/n1) + (1/n2)) = 4.2398

Part b)

Degree of freedom = n1 + n1 - 2 =  14 + 14 - 2 = 26

Part c)

Critical value   t(α/2, n1 + n1 - 2) = t( 0.05/2, 14 + 14 - 2) = 2.056

Part d)

Margin of Error = t(α/2 , n1+n2-2) SP √( (1/n1) + (1/n2)) = 1.2453

Part e)

Confidence interval is :-
( X̅1 - X̅2 ) ± t( α/2 , n1+n2-2) SP √( (1/n1) + (1/n2))
t(α/2, n1 + n1 - 2) = t( 0.05/2, 14 + 14 - 2) = 2.056
( 82.1 - 84.9 ) ± t(0.05/2 , 14 + 14 -2) 1.6025 √ ( (1/14) + (1/14))
Lower Limit = ( 82.1 - 84.9 ) - t(0.05/2 , 14 + 14 -2) 1.6025 √( (1/14) + (1/14))
Lower Limit = -4.0453
Upper Limit = ( 82.1 - 84.9 ) + t(0.05/2 , 14 + 14 -2) 1.6025 √( (1/14) + (1/14))
Upper Limit = -1.5547
95% Confidence Interval is ( -4.0453 , -1.5547 )