Question 1 point
The population standard deviation for the height of college baseball players is 3.0 inches. If we
want to estimate 95% confidence interval for the population mean height of these players with
a 0.68 margin of error, how many randomly selected players must be surveyed? (Round up your
answer to nearest whole number, do not include any decimals) Answer: 75
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Z-Critical Value = NORM.S.INV(.975) = 1.96
n=
32∗1.962.682
3
2
∗
1.96
2
.68
2
Question 2 point
The population standard deviation for the height of college basketball players is 2.9 inches. If we
want to estimate a 99% confidence interval for the population mean height of these players
with a 0.45 margin of error, how many randomly selected players must be surveyed? (Round up
your answer to nearest whole number, do not include any decimals) Answer: 276
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Z-Critical Value = NORM.SINV(.995) = 2.575
n=
2.92∗2.5752.452
2.9
2
∗
2.575
2
.45
2
Question 3 point
The FDA regulates that fresh Albacore tuna fish that is consumed is allowed to contain 0.82 ppm
of mercury or less. A laboratory is estimating the amount of mercury in tuna fish for a new
company and needs to have a margin of error within 0.023 ppm of mercury with 97%
confidence. Assume the population standard deviation is 0.143 ppm of mercury. What sample
size is needed? Round up to the nearest integer, do not include any decimals.
Answer: 183
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Z-Critical Value = NORM.S.INV(.985) = 2.17009
, n=
.1432∗2.170092.0232
.143
2
∗
2.17009
2
.023
2
Question 4 point
A researcher would like to estimate the proportion of all children that have been diagnosed with
Autism Spectrum Disorder (ASD) in their county. They are using 97% confidence level and the
CDC national estimate that 1 in 68 ≈ 0.0147 children are diagnosed with ASD. What sample size
should the researcher use to get a margin of error to be within 1.5%? Round up to the nearest
integer, do not include decimals. Answer 303 correct answer: 304
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Z-Critical Value = NORM.S.INV(.985) = 2.17009
n=
.0147∗.9853∗2.170092.0152
.0147
∗
.9853
∗
2.17009
2
.015
2
Question 5 point
There is no prior information about the proportion of Americans who support gun control in
2019. If we want to estimate 93% confidence interval for the true proportion of Americans who
support gun control in 2019 with a 0.18 margin of error, how many randomly selected
Americans must be surveyed? Answer: (Round up your answer to nearest whole number, do not
include any decimals) 26
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Z-Critical Value = NORM.S.INV(.965) = 1.811911
n=
.5∗.5∗1.8119112.182
.5
∗
.5