CAN SOMEONE HELP ME FIGURE OUT THE "LOWER LIMIT" AND "UPPER LIMIT" ? For a study conducted by the research department of a pharmaceutical company, 215randomly selected individuals were asked to report...

CAN SOMEONE HELP ME FIGURE OUT THE "LOWER LIMIT" AND  "UPPER LIMIT" ?

For a study conducted by the research department of a pharmaceutical company, 215randomly 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 $5.50. A random sample of 295 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 $4.20. 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 90% 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 find the lower limit and upper limit of the

 confidence interval.

CARRY YOUR INTERMEDIATE COMPUTATIONS TO AT LEAST 3 DECIMAL PLACES.ROUND YOUR ANSWER TO AT LEAST 2 DECIMAL PLACES