A study has been made to compare the nicotine contents of two brands of cigarett
ID: 3258158 • Letter: A
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 4.2 milligrams with a standard deviation of 0.7 milligram. Eight cigarettes of Brand B had an average nicotine content of 3.1 milligrams with a standard deviation of 0.5 milligram. Assume that the two sets of data are independent random samples from normal populations with equal variances. Answer the following, and round off your answer to three decimal places. (a) Find a pooled estimate of the population standard deviation. (b) Construct a 95% confidence interval for the difference between the mean nicotine contents of the two brands of cigarettes.Explanation / Answer
Formula:
(x1 bar – x2 bar) (-/+) E
E = tc *SE (x1-x2)
SE (x1-x2) = ((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2) * (1/n1+1/n2)
x1 bar = 4.2
x2 bar = 3.1
s1 = 0.7
s2 = 0.5
n1 = 10
n2 = 8
df = n1 + n2 – 2 = 10 + 8 – 2 = 16
level of significance is 5%
critical tc value is 2.120 from t table
SE = 0.2943
E = 2.120 * 0.2943 = 0.624
x1 bar – x2 bar = 4.2 – 3.1 = 1.1
(x1 bar – x2 bar) (-/+) E
0.476 and 1.724
Question a)
Answer: Pooled estimate of the standard deviation 0.294
Question b)
Answer: (0.476 and 1.724)
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