Homework Answers
Ans:
Standard error of mean=800/sqrt(64)=100
Sample mean=4000
critical z value for 97% Confidence level is 2.17
97% confidence interval for mean annual earnings
=4000+/-2.17*100
=4000+/-217
=(3783, 4217)
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the monthly earnings of a group of business students are normally distribution with a standard devition...
the monthly earnings of a group of business students are
normally distribution with a standard devition of 345 dollars. a
researcher wants to estimate the mean monthly earnings of all
business students. find the sample size needed to have a confidence
level of 89% and a margin of error of 118 dollars.
The monthly earnings of a group of business students are normally distributed with a standard deviation of 345 dollars. A researcher wants to estimate the mean monthly earnings...
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College students annual earnings are normally distributed with standard deviation σ-$800. If the mean earning for gro construct a 95% confidence interval estimate of the mean annual earnings for all college students. 10. up of 64 students is $4000,
the monthly earnings of a group of business students are
normally distribution with a standard devition of 345 dollars. a
researcher wants to estimate the mean monthly earnings of all
business students. find the sample size needed to have a confidence
level of 89% and a margin of error of 118 dollars.
The monthly earnings of a group of business students are normally distributed with a standard deviation of 345 dollars. A researcher wants to estimate the mean monthly earnings...
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