6. (20 pts.) A potential investor conducted a 144-day survey in two theaters to

Question

6. (20 pts.) A potential investor conducted a 144-day survey in two theaters to determine the difference between the average daily attendance. 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 for North Mall is 1,000; while the variance for the South Mall is 1,304. Use the output below to test for a difference in the average daily attendance at the two theaters at a 1% significance level. That is, test Ho: North-Sout 0. Test Statistic = 8 One-sided Critical Value = 2.326 Two-sided critical value = 2.576

Explanation / Answer

Solution:

For the given scenario, we have to use the two sample z test for the population mean because we know the population variances in advance. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: There is no any statistically significant difference in the average daily attendance for North mall theatre and South mall theatre.

Alternative hypothesis: Ha: There is a statistically significant difference in the average daily attendance for North mall theatre and South mall theatre.

H0: µNorth - µSouth = 0 versus Ha: µNorth - µSouth 0

This is a two tailed test.

We are given

Level of significance = = 0.01

Test statistic = Z = 8

Two sided critical value = 2.576

Test statistic value is greater than critical value, so we reject the null hypothesis that there is no any statistically significant difference in the average daily attendance for North mall theatre and South mall theatre.

There is sufficient evidence to conclude that there is a statistically significant difference in the average daily attendance for North mall theatre and South mall theatre.

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