Wednesday
July 30, 2014

Homework Help: Statistics

Posted by Clay on Tuesday, November 27, 2012 at 1:49am.

The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ = 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but measurements on a random sample of 400 cigarettes of this brand gave a mean of x =1.52. Is this evidence that the mean nicotine
content is actually higher than advertised? To answer this, test the hypotheses of
H0: μ = 1.5 vs. Ha: μ > 1.5
at a significance level of α = 0.01.

1. The test statistic for this test is
A) z = -4.00
B) z = -0.20
C) z = 0.20
D) z = 4.00

2. Based on the p-value of the test and the given significance level, what would you
conclude?
A) Fail to reject H0, indicating evidence that the mean nicotine content in this brand of
cigarettes equals 1.5 milligrams.
B) Reject H0, indicating evidence that the mean nicotine content in this brand of
cigarettes is greater than 1.5 milligrams.
C) There is a 5% chance that the null hypothesis is true.
D) We cannot make a conclusion here since we do not know the true mean of the
population.

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Statistics - Im kind of confused on this question?? To help consumers assess the...
Stastics - To help consumers assess the risks they are taking, the Food and Drug...
statistics - A tobacco company claims that the nicotine content of its "light" ...
statistics - A tobacco company claims that the nicotine content of its "light" ...
Statistics - Cigarette companies advertise that the mean amount of nicotine in ...
statistics - How large a sample would be needed to form a 90% confidence ...
statistics - For each of the following questions, decide whether or not the ...
Stor - Here is a simple way to create a random variable X that has mean μ ...
statistics - The mean (μ) of the scale is 98 and the standard deviation (&#...
Statistics - A driving machine was employed to hit a sample of 41 each of two ...

Search
Members