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A 7,000-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 5,500, while the mean attendance on Wednesday evening is calculated as 5,850. 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. Which of the following is the value of the appropriate test statistic?A. z = -2.4736B. z = 2.4736C. tdf=−2.4736D. td f=2.4736

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Answer:

A. z = -2.4736

Explanation:

n1 = 25

n2 = 25

x1 = 5500

x2 = 5850

Sd1= 550

Sd2 = 445

Null hypothesis

μ1 - μ2 = 0

Alternate

μ1 - μ2 != 0

Pooled variance = ((25-1)*550²)+((25-1)*445²)/25+25-2

= 7260000+4752600/48

= 250262.5

Pooled sd = √250262.5

= 500.26243

Test statistic

Z =

5500-5850+10/500.26243*√1/25+1/25

= -2.47357 approximately

-2.4736

A 7,000-seat theater is interested in determining whether there is a difference in-example-1
A 7,000-seat theater is interested in determining whether there is a difference in-example-2
User Abhijeet Khangarot
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