Question

please show all work

2. A study has been made to compare the nicotene contents of two brands of cigarettes. Ten cigarettes of brand A had a mean nicotene content of 3.1 milligrams with a standard deviation of 0.5 milligrams while cight cigarettes of brand B had a mean nicotene content of 2.7 milligrams with a standard deviation of 0.3 milligrams. Assuming that the two sets of data are independent random samples froin a normal populations, construct a 95% confidence interval for the difference between the mean nicotene contents of the two brands of cigarettes. Ans. (-0.00578, 0.80578)

Answer 1

The summary of the statistic are: 2.7 mean std 0.5 0.3 dev 10 The standard error is: 0.5 0.32 10 = 0.1904 The degrees of freedom is

0.5 0.3 10 0.5 0.3 10 10-1 Critical value: The critical value at 5% level of significance at df=15 for a two tailed test is 2.145 (3.1-2.7)+2.145(0.1904) - 0.4000 t 0.40578 (-0.00578,0.80578) Interpretation: One can be 95% confident that the difference between the mean nicotine content of two brands of cigarettes is -0.00578 and 0.80578