Mya applied the distributive property to write the equivalent expressions below
2 answers:
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Answer:
b. We are 95% confident that between 35.5% and 44.3% of the persons in the U.S. labor force is female.
Step-by-step explanation:
1) Data given and notation
n=500 represent the random sample taken
X represent the number of females in the U.S. labor force
estimated proportion of females in the U.S. labor force
represent the significance level (no given, but is assumed)
z would represent the statistic
p= population proportion of females in the U.S. labor force
2) Confidence interval
The confidence interval would be given by this formula
For the 95% confidence interval the value of and
, with that value we can find the quantile required for the interval in the normal standard distribution.
And replacing into the confidence interval formula we got:
On this case the calculated interval is not the given , but let's assume that the confidence interval is given by the statment: "United States Department of Labor find that 95% confidence interval for the proportion of females in the U.S. labor force is .357 to .443."
3) Correct interpretation
a. The margin of error for the true percentage of females in the U.S. labor force is between 35.7% and 44.3%.
False the confidence interval not conclude about the margin of error conclude about the true proportion.
b. We are 95% confident that between 35.5% and 44.3% of the persons in the U.S. labor force is female.
True, the statement report the confidence level and the limits for the margin of error.
c. The percentage of females in the U.S. labro force is between 35.7% and 44.3%.
False, The statement is not correct because not reports the confidence level.
d. All sample of size 500 will yield a percentage of females in the U.S. labor force that falls within 35.7% and 44.3%
False the confidence interval is for the population proportion not just for the samples of size 500
e. None of these.
False, we have an option that is true.