Question

The American Bankers Association reported that, in a sample of 120 consumer purchases in France, 48 were made with cash, compared with 24 in a sample of 55 consumer purchases in the United States.
Construct a 90 percent confidence interval for the difference in proportions. (Round your intermediate value and final answers to 4 decimal places.)

The 90 percent confidence interval is from ___________ to ___________-

Answer 1

Answer:

Step-by-step explanation:

Hello!

You have the information for two variables

X₁: Number of consumer purchases in France that were made with cash, in a sample of 120.

n₁= 120 consumer purchases

x₁= 48 cash purchases

p'₁= 48/120= 0.4

X₂: Number of consumer purchases in the US that were made with cash, in a sample of 55.

n₂= 55 consumer purchases

x₂= 24 cash purchases

p'₂= 24/55= 0.4364

You need to construct a 90% CI for the difference of proportions p₁-p₂

Using the central limit theorem you can approximate the distribution of both sample proportions p'₁ and p'₂ to normal, so the statistic to use to estimate the difference of proportions is an approximate standard normal:

[(p'₁-p'₂) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'_1(1-p'_1)}{n_1} +\frac{p'_2(1-p'_2)}{n_2} }$]

$Z_{0.95}= 1.648$

[(0.4-0.4364)±1.648 * $\sqrt{\frac{0.4(1-0.4)}{120} +\frac{0.4364(1-0.4364)}{55} }$]

[-0.1689;0.0961]

The interval has a negative bond, it is ok, keep in mind that even tough proportions take values between 0 and 1, in this case, the confidence interval estimates the difference between the two proportions. It is valid for one of the bonds or the two bonds of the CI for the difference between population proportions to be negative.

I hope this helps!