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Estimating Risky Behavior 1. We would like to know what percent adult heterosexuals love m e is this mation. The AIDS Behavioral Surveys found that 120 2673 Luther als de tr. Repuding this as a random sample so it's representative of the larger population walt heterosexuals, we want to calculate a 99 confidence for p Pirst calculate D. keeping at least 5 decimal places. Show the conditions for comm e nce Calculate the confidence interval Express that confidence interval in percent, rounding to remove the decimal places for example, 15.2619 to 15 .2 ) Internet the meaning of that goefidence interval ways in the wording of the problem. (3pts.) c. If we generated 4.215 more confidence intervals with the same procedure as above, that is, each time sampling 2.67) heterosexual adults from the population, how many would we expect not to contain p? Round your answer to the nearest whole number (for example, 23.7 to 24).(2pes.) d. Now you are at a crowded adult club filled to maximum capacity. Even though the music is blaring and the multicolored stage lights are flashing, being careful, you take the time to notice where the fire exits are located. By one of them, the Fire Marshall's sign reads, "maximum occupancy 700." Assume the club management is obeying the law and these 700 adults represent a random sample of adult heterosexuals that would have been admissible to our sampling. (Are they really? Just think about it, no need to write out an answer.) Using your confidence interval based on the original sample, estimate the number of adult heterosexuals (as an interval) in the club having multiple partners, rounding to remove the decimal places

Answer 1

(a)

$\hat{p}$ = 170 / 2673 = 0.06360

n$\hat{p}$ = 2673 * 0.06360 = 170

n(1- $\hat{p}$ ) = 2673 * (1 - 0.06360) = 2502.997

Since, n$\hat{p}$ and n(1- $\hat{p}$ ) both are greater than 10, the assumption of normality are met.

(b)

Standard error of sample proportion, SE = $\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{0.0636(1-0.0636)/2673}$ = 0.00472

Z value for 99% confidence interval is 2.576

Margin of error = Z * SE = 2.576 * 0.00472 = 0.0122

99% confidence interval is

(0.06360 - 0.0122, 0.06360 + 0.0122)

(0.0514, 0.0758)

(5%, 8%) (Rounded to nearest percentage)

We're 99% confident that the interval (5%, 8%) captured the true mean percentage of adult heterosexual with multiple partners.

(c)

99% of the intervals will contain p.

Number of confidence interval to contain p = 0.99 * 4215 = 4172.85 $\approx$ 4173

Number of confidence interval to not contain p = 4215 - 4173 = 42

(d)

Using sample size of 700, number of heterosexuals in the club with multiple partners will lie in the range

(700 * 5%, 700 * 8%)

(35, 56)