In: Statistics and Probability
A study reports that 36% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents.
c. The probability is 95% that the sample percentage of Country A companies having three or more female board directors will be contained within what symmetrical limits of the population percentage? The probability is 95% that the sample percentage will be contained above nothing% and below nothing%. (Round to one decimal place as needed.)
Solution
Given that,
p = 36% = 0.36
1 - p = 1 - 0.36 = 0.64
n = 100
= p = 0.36
= [p( 1 - p ) / n] = [(0.36 * 0.64) / 100] = 0.048
Using standard normal table,
P( -z < Z < z) = 95%
= P(Z < z) - P(Z <-z ) = 0.95
= 2P(Z < z) - 1 = 0.95
= 2P(Z < z) = 1 + 0.95
= P(Z < z) = 1.95 / 2
= P(Z < z) = 0.975
= P(Z < 1.96) = 0.975
= z ± 1.96
Using z-score formula,
= z * +
= -1.96 * 0.048 + 0.36
= 0.266
= 26.6%
Using z-score formula,
= z * +
= 1.96 * 0.048 + 0.36
= 0.454
= 45.4%
The probability is 60% that the sample percentage will be contained above 26.6% and below 45.4%