Question

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and non smokers uses a preliminary estimate of the proportion who smoke of .30.a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of.02? use 95% confidence.b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population?c. What is the 95% confidence interval for the proportion of smokers in the population?

Answer 1

a) n = 2017

b) Point estimate is 0.2578

c) 95% Confidence Interval is Minimum = 0.2387, Maximum 0.2769

Explanation:

Here we have

At 95%, we have

`z_(\alpha /2)` = 1.96

To determine sample size, we have

`n = ((z_(\alpha /2))^2 \hat p \hat q)/(E^2)`

Where:

`\hat p` = 0.3

`\hat q` =
`1-\hat p` = 0.7

E = 0.02

Therefore, n = 2016.84 ≈2017

b) The point estimate is given by

`\hat p =(x)/(n) = (520)/(2017)` = 0.2578

c) The confidence interval is given by;

`CI=\hat{p}\pm z* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

Which gives

`CI=0.2578\pm 1.96* \sqrt{(0.2578(1-0.2578))/(2017)}`

Hence CI = Min = 0.2387 to Max = 0.2769

Answer 2

Explanation:

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

a) From the information given,

Margin of error = 0.02

p = 0.3

q = 1 - 0.3 = 0.7

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

α = 1 - 0.5 = 0.05

α/2 = 0.05/2 = 0.025

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

1 - 0.05 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Therefore,

0.02 = 1.96 × √(0.3 × 0.7)/n

0.02/1.96 = √0.21/n

0.0102 = √0.21/n

Taking square of both sides, it becomes

0.00010404 = 0.21/n

n = 0.21/0.00010404

n = 2018

Sample size = 2018

b) if n = 2018

x = 520

Then

p = 520/2018 = 0.26

Point estimate of the proportion of smokers in the population is 0.26

c) q = 1 - 0.26 = 0.74

the 95% confidence interval for the proportion of smokers in the population is

0.26 ± 1.96 × √(0.26)(0.74)/2018

= 0.26 ± 0.019