Question

2. The Carolina Tobacco Company advertised that its best-selling nonfiltered cigarettes contain at most 40 mg of nicotine, but Consumer Advocate magazine ran tests of 10 randomly selected cigarettes and found the amounts (in mg) shown in the list below. It’s a serious matter to charge that the company advertising is wrong, so the magazine editor chooses a significance level of α = 0.01 in testing her belief that the mean nicotine content is greater than 40 mg. Using a 0.01 significance level, test the editor’s belief that the mean is greater than 40 mg. 47.3 39.3 40.3 38.3 46.3 43.3 42.3 49.3 40.3 46.3 Show all the steps of your hypothesis test including a) the claim being tested , b) the null and alternative hypothesis, c) the test statistic, d) the p value, e) initial conclusion, and f) a final conclusion about the original claim.

Answer 1

H₀: = 40

H₁: > 40

= 43.3

s = 3.8006

The test statistic t = ()/(s/)

                           = (43.3 - 40)/(3.8006/)

                            = 2.75

P-value = P(T > 2.75)

           = 1 - P(T < 2.75)

           = 1 - 0.9888

          = 0.0112

Since the P-value is greater than the significance level(0.0112 > 0.01), so we should not reject the null hypothesis.

There is not sufficient evidence to support the claim that the mean nicotine content is greater than 40 mg.