Question 1011623: I really need someones help!
Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.
A simple random sample of 25 filtered 100 mm cigarettes is obtained, and the tar content of each cigarette is measured. The sample has a mean of 19.8 mg and a standard deviation of 3.21 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 100 mm cigarettes is less than 21.1 mg, which is the mean for unfiltered king size cigarettes. What do the results suggest, if anything, about the effectiveness of the filters?
A.
What are the hypotheses?
B.
Identify the test statistic.
t=
(Round to three decimal places as needed.)
Identify the P-value.
The P-value is
(Round to four decimal places as needed.)
C.
State the final conclusion that addresses the original claim. Choose the correct answer below.
D.
What do the results suggest, if anything, about the effectiveness of the filters?
A. The results do not suggest that the filters are effective.
B.The results suggest that the filters increase the tar content.
C.The results suggest that the filters are effective.
D.The results suggest that the filtered cigarettes have the same tar content as unfiltered king size cigarettes.
E.The results are inconclusive because the sample size is less than 30.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a. what are the hypotheses?
the hypotheses are:
null hypothesis is that the mean is greater than or equal to 21.1
alternate hypothesis is that the mean is less than 21.1
b. identity the test statistic.
the test statistic is t = -2.0249
the p-value of the test statistic is p = .0271
c. state the final conclusion.
selection C: the results suggest that the filters are effective.
here's how these conclusions were reached.
the sample mean is 19.8 with a standard deviation of 3.21 and a sample size of 25.
since the standard deviation is from the sample, a t-test is indicted.
the standard error is equal to the standard deviation divided by the square root of the sample size which becomes se = 3.21/5 = .6420.
the formula for t-score is t(dof) = (x-m)/se
x is the mean of your sample = 19.8
m is the mean of the unfiltered cigarettes = 21.1
se is the standard error which is .6420.
dof is the degrees of freedom.
when n = 25, dof = n-1 = 24
your formula becomes t(24) = (19.8-21.1)/.642 which results in:
t(24) = -2.0249
the p-value for a t-score of -2.0249 with 24 degrees of freedom is equal to .02707.
this is less than .05, so the null hypothesis is rejected.
the t-score will tell you the critical t-score which turned out to be -1.7109.
since the t-score of -2.0249 was less than this, the null hypothesis was rejected.
a t-score calculator will tell you the p-value of the t-score.
the online one that i used is at http://stattrek.com/online-calculator/t-distribution.aspx
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