Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 42 who smoke. Suppose a sample of 897 Americans over 42 is drawn. Of these people, 637 don't smoke.

Using the data, construct the 95% confidence interval for the population proportion of Americans over 42 who smoke. Round your answers to three decimal places.

Answer 1

To construct a confidence interval for the population proportion using the given data, we calculate the point estimate and the margin of error. The point estimate is found by dividing the number of successes (those who don't smoke) by the sample size. The margin of error is calculated using the standard error and the critical value. The 95% confidence interval is then determined by adding and subtracting the margin of error from the point estimate.

Step-by-step explanation:

To construct a confidence interval for the population proportion, we will use the formula:

CI = Point Estimate ± Margin of Error

Where:

Point Estimate = Number of successes / Sample size

Margin of Error = Critical value × Standard Error

To find the confidence interval, we first need to calculate the point estimate and the margin of error:

Point Estimate = 1 - (Number of successes / Sample size) = 1 - (637 / 897) = 1 - 0.710 = 0.290

Sample size = 897

Number of successes = 637

Now, let's calculate the margin of error:

Standard Error = sqrt((Point Estimate × (1 - Point Estimate)) / Sample size) = sqrt((0.290 × (1 - 0.290)) / 897) = 0.014

Critical value for a 95% confidence level is approximately 1.96. (You can look up the critical value in a standard normal distribution table or use a calculator.)

Margin of Error = Critical value × Standard Error = 1.96 × 0.014 = 0.027

Now we can construct the confidence interval:

CI = Point Estimate ± Margin of Error = 0.290 ± 0.027

Therefore, the 95% confidence interval for the true proportion of Americans over 42 who smoke is (0.263, 0.317).

Answer 2

The 95% confidence interval for the population proportion of Americans over 42 who smoke is (0.265, 0.325).

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)`.

For this problem, we have that:

897 people, of which 897 - 632 = 265 smoke.

Then
`n = 897, \pi = (265)/(897) = 0.295`

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`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.295 - 1.96\sqrt{(0.295*0.705)/(897)} = 0.265`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.295 - 1.96\sqrt{(0.295*0.705)/(897)} = 0.325`

The 95% confidence interval for the population proportion of Americans over 42 who smoke is (0.265, 0.325).