Question

The american heart association is about to conduct an anti-smoking campaign and wants to know the fraction of americans over 35 35 who smoke. step 2 of 2 : suppose a sample of 632 632 americans over 35 35 is drawn. of these people, 456 456 don't smoke. using the data, construct the 90% 90 % confidence interval for the population proportion of americans over 35 35 who smoke. round your answers to three decimal places?

a) 0.262
b) 0.320
c) 0.278
d) 0.214

Answer 1

To determine the 90% confidence interval for the population proportion of Americans over 35 who smoke, we calculate the sample proportion of smokers, the standard error, and the z-score for the confidence level, and use them to find the margin of error. Then we subtract and add the margin of error to the sample proportion to get the confidence interval.

Step-by-step explanation:

To construct a 90% confidence interval for the population proportion of Americans over 35 who smoke, we can use the data provided with 632 Americans over 35 sampled, of whom 456 do not smoke. This implies that 632 - 456 = 176 do smoke. We start by calculating the sample proportion (p-hat) of smokers as 176/632.

Next, we need the standard error (SE) of the sample proportion, which is given by the formula SE = sqrt((p-hat)*(1-p-hat)/n), where 'n' is the sample size. Then, we determine the z-score associated with a 90% confidence level, which is approximately 1.645 for a two-tailed test.

The margin of error (ME) is then calculated as z-score * SE. Finally, we can construct the confidence interval by adding and subtracting the ME from the sample proportion.

The confidence interval is given by:
(p-hat - ME, p-hat + ME). Using the calculations, we obtain the confidence interval and can round it to three decimal places as requested.