3. A group of researchers want to understand the representation of women in vari

Question

3. A group of researchers want to understand the representation of women in various professions. They start with a poll (generated via a random sample of bar tenders. People currently working in that profession were asked to tick a box identifying themselves as women or leave it blank otherwise. (a) Of the 2100 people in the (random sample, 1176 did not identify as women. How many people did claim to be women, what proportion of the sample was this, and what per- centage of the sample did this represent? (b) What is the standard error of the proportion of women bar-tenders? (c) One of the researchers speculates that the group could half the standard error on the proportion of women by doubling the sample size in future work. Is that likely to be correct? Explain why. (d) Construct a 95% confidence interval around the proportion of women in the study. (e) Construct a 99.992% confidence interval around the proportion of women in the study. (f) In a follow up study, the researchers investigated possible racial biases in promotion. They obtained a random sample of 10 people currently serving as CEOs of Fortune 500 companies. This time the categories into which respondents could place themselves were White', 'Black, 'Other. One respondent identified as Black, seven as White, two as Other. Using the usual formula, what is the standard error on the proportion of CEOs who identify as Black? g) One of the researchers suggests that confidence intervals on the proportion of CEOs who identify as Black are unlikely to be correct if they are based on the usua' formula approximation. Do you agree? Explain why

Explanation / Answer

PART A.
given that,
count of women among the given (x)= 2100 - 1176 = 924
sample size(n)=2100
proportion rate ( p )= x/n = 0.44
PART B.
sample proportion = 0.44
standard error = Sqrt ( (0.44*0.56) /2100) )
= 0.0108

PART C.
True, since sample size is inversely proportional to standard error and it divides
the standard deviation

PART D.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0108
= 0.0212
III.
CI = [ p ± margin of error ]
confidence interval = [0.44 ± 0.0212]
= [ 0.4188 , 0.4612]
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DIRECT METHOD
given that,
possibile chances (x)=924
sample size(n)=2100
success rate ( p )= x/n = 0.44
CI = confidence interval
confidence interval = [ 0.44 ± 1.96 * Sqrt ( (0.44*0.56) /2100) ) ]
= [0.44 - 1.96 * Sqrt ( (0.44*0.56) /2100) , 0.44 + 1.96 * Sqrt ( (0.44*0.56) /2100) ]
= [0.4188 , 0.4612]
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interpretations:
1. We are 95% sure that the interval [ 0.4188 , 0.4612] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

PART E.
confidence interval = [ 0.44 ± 3.944 * Sqrt ( (0.44*0.56) /2100) ) ]
= [0.44 - 3.944 * Sqrt ( (0.44*0.56) /2100) , 0.44 + 3.944 * Sqrt ( (0.44*0.56) /2100) ]
= [0.3973 , 0.4827]

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