Question

Suppose a 90% confidence interval, based upon a sample of size 35, for the mean number of hours that college students spend on social media each week was 25 to 35 hours. The sample standard deviation from this previous study was 17.9. How many students should be surveyed in order to cut the width of the interval down from 10 hours to 5 hours? Group of answer choices A.100 B.300 C.250 D.139

Answer #1

*the 90% confidence
interval is* = (25,35)

so, the margin of error = (35-25)/2 = 5

given data are:-

margin of error (E) = 5/2= 2.5 [ a you want to cut the width of the interval down from 10 hours to 5 hours...i.e, you have to half the margin of error.]

sample standard deviation (s) = 17.9

*z critical value
90% confidence level, both tailed test be:-*

**the needed
sample size be:-**

***in case of doubt, comment below. And if u liked the solution,
please **like.**

Based on a sample of size 49, a 95% confidence interval for the
mean score of all
students, μ, on an aptitude test is from 59.2 to 64.8.
Find the margin of error.
Group of answer choices
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Mean
Sample standard deviation
Population standard deviation
Sample size
Confidence level
Confidence interval
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20
na
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95 %
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20
na
25
90 %
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40
na
25
90 %
100
40
n.a.
16
90 %
100
n.a.
40
16
90 %
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