To construct a 99% confidence interval for the proportion of all adults in the United States who exercise on a regular basis, we can use the formula for a confidence interval for a proportion:
CI = p̂ ± Z(α/2) * √(p̂(1-p̂) / n)
Where:
CI = Confidence interval
p̂ = Sample proportion (85/180 = 0.472)
Z(α/2) = Z-value for a 99% confidence interval (2.576 for a 99% confidence interval)
n = Sample size (180)
Plugging in the values:
CI = 0.472 ± 2.576 * √(0.472*(1-0.472) / 180)
CI = 0.472 ± 2.576 * √(0.24838 / 180)
CI = 0.472 ± 2.576 * 0.030164
CI = 0.472 ± 0.077796
Therefore, the 99% confidence interval for the proportion of all adults in the United States who exercise on a regular basis is (0.394, 0.550) or 0.394 to 0.550 when rounded to three decimal places.
Out of 180
randomly selected adults in the United States who were surveyed, 85
exercise on a regular basis. Construct a 99%
confidence interval for the proportion of all adults in the United States who exercise on a regular basis. Round to three decimal places.
1 answer