Question 826803
For a study conducted by the research department of a pharmaceutical company, 210 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 $17.60 with a standard deviation of $5.40. A random sample of 220 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 $18.40 with a standard deviation of $4.20. 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. 
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.
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sample difference:: 17.60-18.40 = -0.80
ME = 1.96*sqrt[5.4^2/210 + 4.2^2/220] = 0.4748
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What is the lower limit of the 95% confidence interval:-0.8-0.4748 = -1.2748
What is the upper limit of the 95% confidence interval:+0.8+0.4748 = -0.3252
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Cheers,
Stan H.
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