Suppose the nicotine content of two kinds of cigarettes have standard deviations

Question

Suppose the nicotine content of two kinds of cigarettes have standard deviations 1 = 1.2 and 2 = 1.4 milligrams. Fifty cigarettes of the first kind had a sample mean content ¯x1 = 26.1 milligrams, while 40 cigarettes of the second kind had a sample mean content of ¯x2 = 23.8 milligrams.

(a) Do the two kinds of cigarettes have different nicotine content? Answer this question by testing H0 : µ1 = µ2 vs. H1 : µ1 6= µ2 at 0.05 significance level.

(b) Construct a 95% confidence interval for the difference between the two population means µ1 µ2 and make a conclusion.

(c) What is the probability that the interval of part (b) contains the true difference between the two population means µ1 µ2?

(d) If you construct many 95% confidence intervals (as in part(b)), how many of them will contain the true difference µ1 µ2 and how many will not?

Suppose the nicotine content of two kinds of cigarettes have standard deviations 1-1.2 and 2 1.4 milligrams. Fifty cigarettes of the first kind had a sample mean content xi = 26.1 milligrams, while 40 cigarettes of the second kind had a sample mean content of 2- 23.8 milligrams. (a) Do the two kinds of cigarettes have different nicotine content? Answer this question by testing (b) Construct a 95% confidence interval for the difference between the two population means c) What is the probability that the interval of part (b) contains the true difference between the (d) If you construct many 95% confidence intervals (as in part(b)), how many of them will contain Ho : 1 = 2 vs. Hi : 1H2 at 0.05 significance level. 1-2 and make a conclusion. two population means 1-12? the true difference 1-12 and how m any will not?

Explanation / Answer

Here population standard deviation is given.

so here,

1 = 1.2 mg ; 2 = 1.4 mg

so here

standard error of difference between mean nicotine content sed   = sqrt [12 /n1 + 22 /n2] = sqrt [1.22/40 + 1.42/40] = 0.2915

Test statistic

Z = (x1 - x2)/ sed = (26.1 - 23.8)/ 0.2915 = 7.89

so here for two tailed test critical test statistic Z (Critical) = 1.96

Z > Zcritical so we will reject the null hypothesis and conclude that the nicoine content is higher in Cigerette 1 than cigrette 2

(b) 95% confidence interval = (x1 - x2) +- Z95% sed = (26.1 - 23.8) +- 1.96 * 0.2915 = (1.729, 2.871)

here we see tha 95% confidence interval doesn't consists the value of zero in the confidence interval so the mean nicotine content is different for both type of cigrettes

(c) Here there is 95% probability that the interval of part (b) contains the true difference between the two population means 1 - 2

(d) Here there are 95% confidence interval contains the true difference 1 - 2 and 5% will not contain the true difference.

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