Question

On a survey taken in 2018 and reported in the GSS data​ file, the​ question, "How much of your communication is via​ text, mobile​ phone, or​ internet?", was asked.

The summary results are as​ follows: Among 202 people who are​ 18-29 years​ old, 69​ reported, "All or almost all of​ it." Among 166 people who are​ 40-49 years​ old, 50​ reported, "All or almost all of​ it."

a. Find the​ 97% confidence interval for the difference between the proportions of those two age groups that​ answerd, "All or almost all of​ it." Include the name of the routine and the margin of error in your answer.

b. Does this data indicate that the population proportion of this issue is different between the two​ groups? Explain.

Answer 1

We want to know about the population proportion difference. That means these are binomial proportion differences. Since n > 30, we can use normal approximation.

Here the success is to answer yes to almost or all of it.

  

	18-29 (1)	40-49 (2)
x	69	50
n	202	166
Sample	0.3416	0.3012
pooled	0.3234

a. Find the​ 97% confidence interval for the difference between the proportions of those two age groups that​ answerd, "All or almost all of​ it." Include the name of the routine and the margin of error in your answer.

   =1 - 0.97 = 0.03

Therefore the critical value at

=

= 2.17009.............using normal percentage tables with p = 0.015

Margin of error = Critical value * SE

= *

= 2.17009 * 0.0024

Subsituting the values

This means that we are 97% confident that true proportion difference lies within this range.

b. Does this data indicate that the population proportion of this issue is different between the two​ groups? Explain.

We can answer this using confidence interval testing for difference.

Test:

Null: There is no difference

Alternative: There is a difference

Test Statistic:

(0.038, 0.0428)

Since the interval does not include '0', we can say that there is a difference between the proportions at 3% significance.