Is this the correct way for finding the margin of error of two samples?

Question

What I have for (b) is

1.96(sqrt(207.11/8.00 + 200.45/6.10)) = 15.023

Something seems off. Not sure about the second part.

Best will surely be upvoted.

Thanks

Answer 1

Here is Minitab output for a Welch (separate variances) 95% t confidence interval $(3.78, 9.54):$

Two-Sample T-Test and CI 

Sample   N    Mean  StDev  SE Mean
1       50  207.11   8.00      1.1
2       45  200.45   6.10     0.91

Difference = μ (1) - μ (2)
Estimate for difference:  6.66
95% CI for difference:  (3.78, 9.54)
T-Test of difference = 0 (vs ≠): T-Value = 4.59  P-Value = 0.000  DF = 90

Notice that the CI does not contain 0, so it seems men plan to spend significantly more than women. The point estimate of the difference is $6.66.

Perhaps you are supposed to do a pooled confidence interval. Minitab's confidence interval of this type is: $(3.74, 9.58).$ This procedure assumes that the population variances for men an women are the same and uses the pooled estimate $S_p = 7.1642$ of the population standard deviation.

The Welch procedure is preferred by many practicing statisticians, but the the pooled procedure is still presented in many textbooks (possibly because it is a little easier to compute for homework and exams).

You should look in your text or notes to find the formulas for getting these confidence intervals, then do the appropriate computation and see that it matches the Minitab output.