The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation a = 0.1. The brand advertises that the mean nicotine content of their Cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised, He takes measurements from a SRS of 15 cigarettes of this brand. The sample yields an average of 1.45 mg of nicotine. Conduct a test using a significance level of a 0.05. (a) The test statistic (b) The critical value, z* (c) The final conclusion is A. The nicotine content is probably higher than advertised. B. There is not sufficient evidence show that the ad is misleading.

MATLAB: An Introduction with Applications
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Author:Amos Gilat
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The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o =
0.1. The brand advertises that the mean nicotine content of their
cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 15 cigarettes of this brand. The
sample yields an average of 1.45 mg of nicotine. Conduct a test using a significance level of a =
0.05.
(a) The test statistic
(b) The critical value, z* =
(c) The final conclusion is
A. The nicotine content is probably higher than advertised.
B. There is not sufficient evidence to show that the ad is misleading.
Transcribed Image Text:The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o = 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 15 cigarettes of this brand. The sample yields an average of 1.45 mg of nicotine. Conduct a test using a significance level of a = 0.05. (a) The test statistic (b) The critical value, z* = (c) The final conclusion is A. The nicotine content is probably higher than advertised. B. There is not sufficient evidence to show that the ad is misleading.
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