Question

A recent Gallup poll found that 36% of U.S. teens aged from 13 to 17 years old have a computer with [(6)] Internet access in their rooms. The poll was based on a random sample of 1028 teens. (a) In words, determine the population parameter, p, that this sample is estimating. [1] (b) Produce a 99% confidence interval for the population proportion, p, and interpret what it means in [3] this context. Round your answers to three decimals. (c) How big of a sample must we take in order to ensure that we are within 1% of the true proportion [2] with 99% confidence

Answer 1

Explanation:

a) The population parameter is the population proportion.

b) Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 1028

p = 36/100 = 0.36

q = 1 - 0.36 = 0.64

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.99 = 0.1

α/2 = 0.01/2 = 0.005

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.05 = 0.995

The z score corresponding to the area on the z table is 2.58. Thus, confidence level of 99% is 2.58

Therefore, the 99% confidence interval is

0.36 ± 2.58 × √(0.36)(0.64)/1028

The lower limit of the confidence interval is

0.36 - 0.039 = 0.321

The upper limit of the confidence interval is

0.36 + 0.039 = 0.399

Therefore, with 99% confidence interval, the proportion of U.S. teens aged from 13 to 17 years old that have a computer with Internet access in their rooms is between 0.321 and 0.399

c) for a margin of error of 1%, that is 1/100 = 0.01, then

0.01 = 2.58 × √(0.36)(0.64)/n

0.01/2.58 = √0.2304/n

0.00387596899 = √0.2304/n

Square both sides

0.00001502314 = 0.2304/n

n = 0.2304/0.00001502314

n = 15336