Hello This is my 4th attempt at getting a confidence interva

Hello,

This is my 4th attempt at getting a confidence interval for this problem. So I hope the tutor is actually reading this. Please see the table below. I need the unanswered boxes answers, then a Confidence Interval Determined for the sum of two variances?

Mean of Public

27.5

Mean of private

26.5

Standard Deviation of public

606.194

Standard Deviation of private

507.35

Count of public

26

Count of private

26

Difference in mean[RB1]

4.6278

Standard Deviation of public squared

24.62

Standard Deviation of private squared

22.52

Variance (public)/count __________

Variance (private)/count

________________

Sum of standard deviation values

Square root result of previous formula

_________

z value for 95% confidence interval:___________

Margin of error: _________

Upper Limit:_________

Lower Limit:_______

Due to the difference in variances, we find that the confidence interval of 95% for this data is ______?

Solution

Given SD (public) = 24.62 so Var(public) = 606.194 and SD(private) = 22.52 so Var(private) = 507.35

Variance (public)/count = (606.194) / 26 = 23.3151

Variance (private)/count = 507.35 / 26 = 19.5135

Sum of standard deviation values = 24.62+22.52 = 47.14 (but it is not required for further process) So i calculated

as 23.3151 + 19.5135 = 42.8286

Square root result of previous formula = 6.5444

z value for 95% confidence interval: 1.96

Margin of error: 1.96* 6.5444 = 12.827

Upper Limit: 4.6278 + 12.827 = 17.4548

Lower Limit: 4.6278 - 12.827 = -8.1992

The confidence interval of 95% for the data is (-8.1992, 17.4548)