A small town claims that it needs rezoning to accommodate its current growth.The town wants to re-designate a current residential area as commercial.To gain approval from the zoning commission, it must show that the commercial taxes paid in this area are significantly higher, on average, than those paid in other residentially zoned areas.A random sample of 16 dwellings in standard residential areas (sample 1) yields an average commercial tax of $25,000 with a standard deviation of $11,000.A sample of 15 dwellings in the "rezoned" area (sample 2) has an average commercial tax of $55,800 with a standard deviation of $2500.Assume that the populations in question are both normal and have unequal variance.Construct a 90% confidence interval on the population mean difference in commercial taxes between the two areas.
A) (-35,882.27, -26,011.70)
B) (-35,588.26, -26,011.70)
C) (-38,320.51, -23,279.50)
D) (-35,746.04, -25,854.00)

Question

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is (-35,882.27, -26,011.70). This interval is calculated using the formula:CI = (x1 - x2) +/- t*sqrt(s1^2/n1 + s2^2/n2)where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and t is the t-value for the desired confidence level and degrees of freedom.In this case, x1 = 25,000, x2 = 55,800, s1 = 11,000, s2 = 2500, n1 = 16, n2 = 15, and t = 2.131 (for a 90% confidence level and 29 degrees of freedom).Plugging these values into the formula, we get:CI = (25,000 - 55,800) +/- 2.131*sqrt(11,000^2/16 + 2500^2/15)CI = (-35,882.27, -26,011.70)

A Small Town Claims That It Needs Rezoning to Accommodate