# How to calculate the 95% confidence interval for this problem?

More and more college students are choosing not to buy the meal plan on campus and instead buy or even make their own meals. A study at a local university revealed that in a random sample of 250 seniors last year, 49% of them no longer bought the meal plan, whereas this year, in a different random sample of 240 seniors, 62% of them no longer bought the meal plan.

What is the 95% confidence interval for the true difference in proportions of seniors between this year and last year who do not buy the meal plan anymore? (Round your answers to three decimal places.)

(____,____)

I figured out the point estimate to be: 0.13

And the standard error to be: 0.04451

and the margin of error with 95% confidence to be: 0.087

But having difficulty finding that interval.

Thank you

You have all the necessary ingredients. A $100(1-\alpha)$% confidence interval is given by

$$\hat{p}_1-\hat{p}_2\pm z_{\alpha/2}\cdot s.e.(\hat{p}_1-\hat{p}_2)$$

where $z_{0.025} \approx 1.96$

and

\begin{align*} s.e.(\hat{p}_1-\hat{p}_2) &=\sqrt{\frac{\hat{p}_1(1−\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1−\hat{p}_2)}{n_2}}\\\\ &=\sqrt{\frac{0.62\cdot0.38}{240}+\frac{0.49\cdot0.51}{250}}\\\\ &\approx0.04451 \end{align*}

You have correctly obtained the point estimate of $0.13$ and the margin of error to be $0.08724$

A $95$% confidence interval would thus be $(0.043,0.217)$

• Thanks. Don't understand where the 0.043, and 0.217. Could you explain where it came from? – Soscrates Fd Apr 3 '18 at 14:22

I wonder who makes up these problems with contradictory percentages and sample sizes: 49% of 250 is 122.5 students. Is someone undecided? (In my experience very few students are undecided about college meal plans.)

Anyhow, approximate integer values entered into Minitab 17 gave the following result:

Sample    X    N  Sample p
1       123  250  0.492000
2       149  240  0.620833

Difference = p (1) - p (2)
Estimate for difference:  -0.128833
95% CI for difference:  (-0.216059, -0.0416078)
Test for difference = 0 (vs ≠ 0):  Z = -2.89  P-Value = 0.004


The 95% CI $(-.216, -.042)$ or (reversing the years) $(.042, .216)$ does not include $0,$ so there seems to have been a change from one year to the next. Allowing for the rounding (?) difficulties mentioned above, this is the same as @Remy's Answer. (+1)

Note about fake data. Old adage among practicing statisticians: "67.53% of all statistics are made up on the spot."