Question

In a survey of 3500 adults (18 and over) who reside in California, 2339 were found to like outdoor sports. Construct a confidence estimate for the true proportion of adults who live in California and who like outdoor sports. Use a confidence level of 99%. Write your answer exactly as is given by Excel. Do not round.

a) =CONFIDENCE.T(0.01, 2339, 3500)
b) =CONFIDENCE.NORM(0.01, 2339, 3500)
c) =CONFIDENCE.T(0.01, 2339, 3500, 2)
d) =CONFIDENCE.NORM(0.01, 2339, 3500, 2)

Answer 1

To construct a confidence interval for the true proportion of adults who live in California and like outdoor sports, we can use the formula: CI = p ± Z * sqrt((p * (1-p)) / n), where p is the proportion of adults who like outdoor sports, Z is the Z-score corresponding to the desired confidence level, and n is the sample size. Given that 2339 out of 3500 adults surveyed like outdoor sports, the confidence interval for the true proportion is [0.6404, 0.6956].

Step-by-step explanation:

To construct a confidence interval for the true proportion of adults who live in California and like outdoor sports, we can use the formula:

CI = p ± Z * sqrt((p * (1-p)) / n)

where:

• CI is the confidence interval
• p is the proportion of adults who like outdoor sports
• Z is the Z-score corresponding to the desired confidence level
• sqrt is the square root function
• n is the sample size

Given that 2339 out of 3500 adults surveyed like outdoor sports, the point estimate for the proportion is p = 2339/3500 = 0.668. The Z-score corresponding to a 99% confidence level is approximately 2.576. Substituting these values into the formula, we get:

CI = 0.668 ± 2.576 * sqrt((0.668 * (1-0.668)) / 3500)

Calculating this expression gives us:

CI = 0.668 ± 2.576 * 0.0107

Simplifying further, we find:

CI = 0.668 ± 0.0276

Therefore, the confidence interval for the true proportion of adults who live in California and like outdoor sports is [0.6404, 0.6956].