Asked by Roger
To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette yielded mean nicotine content of 28.8 milligrams and standard deviation of 2.1 milligrams for a sample of cigarettes. Construct a 90% confidence interval for the mean nicotine content of this brand of cigarette.
Choose one answer. A. 28.8 ± 1.361
B. 28.8 ± 1.381
C. 28.8 ± 1.283
D. 28.8 ± 1.302
Choose one answer. A. 28.8 ± 1.361
B. 28.8 ± 1.381
C. 28.8 ± 1.283
D. 28.8 ± 1.302
Answers
Answered by
PsyDAG
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z value for the smaller portion = .05. (90% confidence interval leaves 5% at each end.)
Mean ± 1.645(SD) = ?
However, this does not match any of the answers given. Do you have any typos?
Mean ± 1.645(SD) = ?
However, this does not match any of the answers given. Do you have any typos?
Answered by
jc
no, you have to use a t value. you did the problem wrong.
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