Question

The Statistical Abstract of the United States reported the percentage of people age 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 0.12. How large a sample should be taken to estimate the actual proportion of smokers with a margin of error of 0.02

Answer 1

To estimate the actual proportion of smokers with a margin of error of 0.02, you need to determine the sample size using the formula n = (Z*Z * p * (1-p)) / E*E. For a preliminary estimate of 0.12 and a margin of error of 0.02, the sample size should be 1011.

Step-by-step explanation:

To estimate the actual proportion of smokers with a margin of error of 0.02, you need to determine the sample size. You can use the formula:

n = (Z*Z * p * (1-p)) / E*E

Where:

• n = sample size
• Z = Z-score corresponding to the desired confidence level
• p = preliminary estimate of the proportion who smoke
• E = margin of error

Substituting the given values into the formula:

n = (Z*Z * p * (1-p)) / E*E = (Z*Z * 0.12 * 0.88) / 0.02*0.02

Since the confidence level is not specified, let's assume a 95% confidence level, which corresponds to a Z-score of approximately 1.96. Substituting this value into the formula:

n = (1.96*1.96 * 0.12 * 0.88) / 0.02*0.02

Simplifying the equation:

n = (3.8416 * 0.1056) / 0.0004 = 1010.6

Rounding up to the nearest whole number, the sample size should be 1011.

Answer 2

A sample of 1015 should be taken.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is of:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 0.12.

This means that
`\pi = 0.12`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`. I used 95% because the confidence level was not given.

How large a sample should be taken to estimate the actual proportion of smokers with a margin of error of 0.02?

This is n for which M = 0.02. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.02 = 1.96\sqrt{(0.12*0.88)/(n)}`

`0.02√(n) = 1.96√(0.12*0.88)`

`√(n) = (1.96√(0.12*0.88))/(0.02)`

`(√(n))^2 = ((1.96√(0.12*0.88))/(0.02))^2`

`n = 1014.2`

Rounding up:

A sample of 1015 should be taken.