# Question: With reference to Example 13 4 for what values of x1

With reference to Example 13.4, for what values of x1 – x2 would the null hypothesis have been rejected? Also find the probabilities of type II errors with the given criterion if

(a) µ1 – µ2 = 0.12;

(b) µ1 – µ2 = 0.16;

(c) µ1 – µ2 = 0.24;

(d) µ1 – µ2 = 0.28.

Example 13.4

An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If n1 = 50 cigarettes of the first kind had an average nicotine content of 1 = 2.61 milligrams with a standard deviation of s1 = 0.12 milligram, whereas n2 = 40 cigarettes of the other kind had an average nicotine content of 2 = 2.38 milligrams with a standard deviation of s2 = 0.14 milligram, test the null hypothesis µ1 – µ2 = 0.20 against the alternative hypothesis µ1 – µ2 ≠ 0.20 at the 0.05 level of significance. Base the decision on the P– value corresponding to the value of the appropriate test statistic.

(a) µ1 – µ2 = 0.12;

(b) µ1 – µ2 = 0.16;

(c) µ1 – µ2 = 0.24;

(d) µ1 – µ2 = 0.28.

Example 13.4

An experiment is performed to determine whether the average nicotine content of one kind of cigarette exceeds that of another kind by 0.20 milligram. If n1 = 50 cigarettes of the first kind had an average nicotine content of 1 = 2.61 milligrams with a standard deviation of s1 = 0.12 milligram, whereas n2 = 40 cigarettes of the other kind had an average nicotine content of 2 = 2.38 milligrams with a standard deviation of s2 = 0.14 milligram, test the null hypothesis µ1 – µ2 = 0.20 against the alternative hypothesis µ1 – µ2 ≠ 0.20 at the 0.05 level of significance. Base the decision on the P– value corresponding to the value of the appropriate test statistic.

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