.4.15-T Assume that a simple random sample has been selected from a nomally dist
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Question
.4.15-T Assume that a simple random sample has been selected from a nomally distributed population and test the given claim. ldenify the null and alternative hypotheses, test statibtic, P-value and state the final concusion that addesses the original claim A simple sandom sample of 25 fihered 100mm cigarettes is obtained, and the tar content of each cigarethe is measured The sample has a mean of 18 8 mg and a standard devlation of 381 mg Use a 0 05 signicance level to test the claim that the mean tar content of Shered 100 mm cigarettes is less than 21 1 mg which is the mean for undilbered king size dgaretes What do the resuts suggest Yanything about the effectiveness of the hers? What are the hypotheses? H1 ?Explanation / Answer
Given :-
sample mean is X¯=18.8 and
sample standard deviation is s=3.81, and sample size is n=25.
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: ? = 21.1
Ha: ? < 21.1
This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.
(2) Rejection Region
The significance level is ?=0.05, and the critical value for a left-tailed test is tc?=?1.711.
The rejection region for this left-tailed test is R = {t: t < -1.711}
(3) Test Statistics
The t-statistic is computed as follows:
t = [ (?X¯??0??) / (s/?n) ] = [ (?18.8?21.1) / (3.81/?25) ]? = ?3.018
(4) Decision about the null hypothesis
Since it is observed that t=?3.018 < tc?=?1.711, it is then concluded that the null hypothesis is rejected.
P-value : The p-value is p = 0.003 and since p = 0.003 < 0.05, it is concluded that the null hypothesis is rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean ? is less than 21.1 at the 0.05 significance level.
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