For a study conducted by the research department of a pharmaceutical company, 28
ID: 3153662 • Letter: F
Question
For a study conducted by the research department of a pharmaceutical company, 285 randomly selected individuals were asked to report the amount of money they spend annually on prescription allergy relief medication. The sample mean was found to be $17.70 with a standard deviation of $5.70. A random sample of 290 individuals was selected independently of the first sample. These individuals reported their annual spending on non-prescription allergy relief medication. The mean of the second sample was found to be $18.30 with a standard deviation of $4.10. As the sample sizes were quite large, it was assumed that the respective population standard deviations of the spending for prescription and non-prescription allergy relief medication could be estimated as the respective sample standard deviation values given above. Construct a 95% confidence interval for the difference µ1 - µ2 between the mean spending on prescription allergy relief medication (µ1)) and the mean spending on non-prescription allergy relief medication (µ2)). Then complete the table below.
• What is the lower limit of the 95% confidence interval?
• What is the upper limit of the 95% confidence interval?
Explanation / Answer
Calculating the means of each group,
X1 = 17.7
X2 = 18.3
Calculating the standard deviations of each group,
s1 = 5.7
s2 = 4.1
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 285
n2 = sample size of group 2 = 290
Also, sD = 0.414687252
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
Z(alpha/2) = 1.959963985
Hence,
lower bound = [X1 - X2] - z(alpha/2) * sD = -1.412772079 [ANSWER]
upper bound = [X1 - X2] + z(alpha/2) * sD = 0.212772079 [ANSWER]
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