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Chapter 10 Statistical Inference About Means And Proportions With Two Populations

3502 words - 15 pages

Chapter 10
Statistical Inference About Means and Proportions with Two Populations


Learning Objectives

1. Be able to develop interval estimates and conduct hypothesis tests about the difference between two population means whenandare known.

2. Know the properties of the sampling distribution of .

3. Be able to use the t distribution to conduct statistical inferences about the difference between two population means whenandare unknown.

4. Learn how to analyze the ...view middle of the document...

1.96



2  1.17 (.83 to 3.17)

2. a.

b. p-value = 1.0000 - .9788 = .0212

c. p-value .05, reject H0.

3. a.

b. p-value = 2(.0630) = .1260

c. p-value > .05, do not reject H0.

4. a. = population mean for smaller cruise ships

= population mean for larger cruise ships

= 85.36 – 81.40 = 3.96

b.



c. 3.96 ± 1.88 (2.08 to 5.84)

5. a. = 135.67 – 68.64 = 67.03

b.

c. 67.03  17.08 (49.95 to 84.11) We estimate that men spend $67.03 more
than women on Valentine’s Day with a margin of error of $17.08.

6. = mean hotel price in Atlanta

= mean hotel price in Houston

H0:
Ha:



p-value = .0351

p-value .05; reject H0. The mean price of a hotel room in Atlanta is lower than the mean price of a hotel room in Houston.

7. a. = population mean satisfaction score for Target customers

= population mean satisfaction score for Walmart customers

H0:
Ha:

b. = 79 – 71 = 8



For this two-tailed test, p-value is two times the upper-tail area at z = 2.46.

p-value = 2(1.0000 – .9931) = .0138

p-value .05; reject H0. The population mean satisfaction scores differ for the two retailers.

c.



8 6.37 (1.63 to 14.37)

Target shows a higher population mean customer satisfaction score than Walmart with the 95% confidence interval indicating that Target has a population mean customer satisfaction score that is 1.63 to 14.37 higher than Walmart.

8. a. This is an upper tail hypothesis test.
...

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