A potential investor conducted a 144 day survey in each theater in order to dete

Question

A potential investor conducted a 144 day survey in each theater in order to determine the difference between the average daily attendance at the North Mall and South Mall theaters. The North Mall Theater averaged 630 patrons per day; while the South Mall Theater averaged 598 patrons per day. From past information, it is known that the variance in attendance for North Mall is 1,000; while the variance for the South Mall is 1,304. Develop a 95% confidence interval for the difference between the average daily attendance at the two theaters. Interpret your results. (if you use the TI83/84, state the correct z or t value as well).

Explanation / Answer

Let N be the mean attendance at theater North and S at theater south. We want some information on the mean and variance of N-S. Since we presume they are independent, the variance of N?S is Var[N]+Var[S]. The mean is the difference of the observed means.

Variance of N-S is (1000+1304) / 144 because variance of the mean is divided by the sample size. Since the variance is not computed based off the sample data, we would use N rather than N?1 as the divisor. Also, since the exercise gives N as a perfect square, that's usually a hint it's safe to use N as the divisor :-)

Var[N?S]=2304/144=16, so ?=4

Assuming N?S is approximately normal with mean 32 and ?=4, the 95% confidence interval is given by 32?1.96*4 to 32+1.96*4 which is [24.16 , 39.84]



N S Diff
Mean 630 598 32
Var 1000 1304 2304
Var of Mean 16
Std Err 4

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