Question

The Millennial generation (so called because they were born after 1980 and began to come of age around the year 2000) is less religiously active than older Americans. One of the questions in the General Social Survey in 2010 was "How often does the respondent pray?" Among the 419 respondents in the survey between 18 and 30 years of age, 277 prayed at least once a week. Assume that the sample is an SRS. Use the plus-four method to give a 99% confidence interval (±0.0001) for the proportion of all adults between 18 and 30 years of age who pray at least once a week. 99% confidence interval is from _ to _

Answer 1

Solution :

Given that,

n = 419

x = 277

Point estimate = sample proportion = = x / n = 0.6611

1 - = 0.3389

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.6611 * 0.3389) / 419)

= 0.0596

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.6611 - 0.0596 < p < 0.6611 + 0.0596

0.6015 < p < 0.7207

The 99% confidence interval form 0.6015 to 0.7207