Thursday

March 5, 2015

March 5, 2015

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

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.

- Statistics -
**PsyDAG**, Tuesday, November 27, 2012 at 12:56pmZ = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score to answer 2.

**Answer this Question**

**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 ...