(a)
= 170 / 2673 = 0.06360
n = 2673 * 0.06360 = 170
n(1- ) = 2673 * (1 - 0.06360) = 2502.997
Since, n and n(1- ) both are greater than 10, the assumption of normality are met.
(b)
Standard error of sample proportion, SE = = 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 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)
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