A 7000-seat theater is interested in determining whether there is a difference i

Question

A 7000-seat theater is interested in determining whether there is a difference in attendance between shows on Tuesday evening and those on Wednesday evening. Two independent samples of 25 weeks are collected for Tuesday and Wednesday. The mean attendance on Tuesday evening is calculated as 5500, while the mean attendance on Wednesday evening is calculated as 5850. The known population standard deviation for attendance on Tuesday evening is 550 and the known population standard deviation for attendance on Wednesday evening is 445.

Refer to Exhibit 10.4. What is the appropriate decision given a 5% level of significance?

Conclude that the mean attendance differs since the p-value = 0.0067 < 0.05.

Conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

Do not conclude that the mean attendance differs since the p-value = 0.0067 < 0.05.

Do not conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

Conclude that the mean attendance differs since the p-value = 0.0067 < 0.05.

Conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

Do not conclude that the mean attendance differs since the p-value = 0.0067 < 0.05.

Do not conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

Explanation / Answer

Given that,

mean(x)=5500

standard deviation , 1 =550

number(n1)=25

y(mean)=5850

standard deviation, 2 =445

number(n2)=25

null, Ho: u1 = u2

alternate, H1: 1 != u2

level of significance, = 0.05

from standard normal table, two tailed z /2 =1.96

since our test is two-tailed

reject Ho, if zo < -1.96 OR if zo > 1.96

we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)

zo=5500-5850/sqrt((302500/25)+(198025/25))

zo =-2.4736

| zo | =2.4736

critical value

the value of |z | at los 0.05% is 1.96

we got |zo | =2.474 & | z | =1.96

make decision

hence value of | zo | > | z | and here we reject Ho

P-Value : Two Tailed ( double the one tail ) - Ha = 0.0134

hence value of p0.05 > 0,here we reject Ho

ANSWERS

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Conclude that the mean attendance differs since the p-value = 0.0134 < 0.05.

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