Chapter 10 hypothesis testing
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- QUESTIONS 70 THROUGH 73 ARE BASED ON THE FOLLOWING INFORMATION
- QUESTIONS 78 THROUGH 82 ARE BASED ON THE FOLLOWING INFORMATION
- Summary statistics for two samples
- Test of difference=0
- ANSWER
69. A recent study of educational levels of 1000 voters and their political party affiliations in a Midwestern state showed the results given in the table below. Use the 5% significance level and test to determine if party affiliation is independent of the educational level of the voters.
ANSWER:  We fail to reject the null hypothesis of independence at the 5% significance level (since p-value = 0.087 > .05). We may conclude that party affiliation is independent of the educational level of the voters. QUESTIONS 70 THROUGH 73 ARE BASED ON THE FOLLOWING INFORMATION: A marketing research consultant hired by Coca-Cola is interested in determining if the proportion of customers who prefer Coke to other brands is over 50%. A random sample of 200 consumers was selected from the market under investigation, 55% favored Coca-Cola over other brands. Additional information is presented below.
70. If you were to conduct a hypothesis test to determine if greater than 50% of customers prefer Coca-Cola to other brands, would you conduct a one-tail or a two-tail hypothesis test? Explain your answer. ANSWER: One-tailed, since the consultant is interested in finding whether the proportion is actually greater than 50%. 71. How many customers out of the 200 sampled must have favored Coke in this case? ANSWER: (200)(0.55) = 110 72. Using a 5% significance level, can the marketing consultant conclude that the proportion of customers who prefer Coca-Cola exceeds 50%? Explain your answer. ANSWER: No. You cannot reject the null hypothesis at a 5% level of significance, since p-value = 0.07761 > 0.05. 73. If you were to use a 1% significance level, would the conclusion from part c change? Explain your answer. ANSWER: No. You still cannot reject the null hypothesis at a 1% level of significance, since p-value = 0.07761 > 0.01. QUESTIONS 74 THROUGH 77 ARE BASED ON THE FOLLOWING INFORMATION: The owner of a popular Internet-based auction site believes that more than half of the people who sell items on her site are women. To test this hypothesis, the owner sampled 1000 customers who sale items on her site and she found that 53% of the customers sampled were women. Some calculations are shown in the table below
74. If you were to conduct a hypothesis test to determine if greater than 50% of customers who use this Internet-based site are women, would you conduct a one-tail or a two-tail hypothesis test? Explain your answer. ANSWER: One-tailed, since the owner is interested in finding whether the proportion is actually greater than 50%. 75. How many customers out of the 1000 sampled must have been women in this case? ANSWER: (1000)(0.53) = 530 76. Using a 5% significance level, can the owner of this site conclude that women make up more than 50% of her customers? Explain your answer. ANSWER: Yes. You can reject the null hypothesis at a 5% level of significance, since p-value = 0.0287 < 0.05. 77. If you were to use a 1% significance level, would the conclusion from Question 76 change? Explain your answer. ANSWER: Yes. Your answer would now change. You cannot reject the null hypothesis at a 1% level of significance, since p-value = 0.0287 > 0.01. QUESTIONS 78 THROUGH 82 ARE BASED ON THE FOLLOWING INFORMATION: Q-Mart is interested in comparing customer who used it own charge card with those who use other types of credit cards. Q-Mart would like to know if customers who use the Q-Mart card spend more money per visit, on average, than customers who use some other type of credit card. They have collected information on a random sample of 38 charge customers and the data is presented below. On average, the person using a Q-Mart card spends $192.81 per visit and customers using another type of card spend $104.47 per visit.
78. Given the information above, what is ANSWER:  . This represents a one-tail test.
79. What are the degrees of freedom for the t-statistic in this calculation? Explain how you would calculate the degrees of freedom in this case. ANSWER: d.f = 13 + 25 – 2 = 36 80. What is the assumption in this case that allows you to use the pooled standard deviation for this confidence interval? ANSWER: The assumption is that the two populations standard deviations are equal; that is 81. Using a 5% level of significance, is there sufficient evidence for Q-Mart to conclude that customers who use the Q-Mart card charge, on average, more than those who use another charge card? Explain your answer. ANSWER: Yes. There is a statistical difference between those using the Q-Mart card and those who use other types of charge cards, since p-value = 0.006 < 0.05. 82. Using a 1% level of significance, is there sufficient evidence for Q-Mart to conclude that customers who use the Q-Mart card charge, on average, more than those who use another charge card? Explain your answer. ANSWER: Yes. There is still a statistical difference between those using the Q-Mart card and those who use other types of charge cards, since p-value = 0.006 < 0.01. Download 360.93 Kb. Share with your friends:
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for this comparison? Also, does this represent a one-tailed or a two-tailed test? Explain your answer.