Question

A study has been made to compare the nicotine contents of two brands of cigarettes. Ten cigarettes of Brand A had an average nicotine content of 3.1 milligrams with a standard deviation of of 0.5 milligram, while 8 cigarettes of Brand B had an average nicotine content of 2.7 milligrams with a standard deviation of 0.7 milligram. Assuming that the two sets of data are independent random samples from normal populations with equal variances, construct a 95% confidence interval for the difference between the mean nicotine contents of the two brands of cigarettes.

Answer 1

−0.2226 ; 1.0226

Explanation:

Given :

n1 = 10 ; n2 = 8 ; s1 = 0.5 ; s2 = 0.7 ; x1 = 3.1 ; x2 = 2.7

Confidence interval, two - sample, t procedure ;

μ1 - μ2 = (x1 - x2) ± Tcritical*S

S = sqrt[(s1²/n1) + (s2²/n2)]

S = sqrt[(0.5^2/10) + (0.7^2/8)]

S = sqrt[0.08625]

S = 0.2937

Tcritical at 95% confidence interval

df = 10+8 - 2= 16

Tcritical = 2.12

(3.1 - 2.7) ± 2.12*(0.2937)

0.4 ± 0.6226

(0.4 - 0.6226) ; (0.4 + 0.6226)

−0.2226 ; 1.0226