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Marketing Research Sample Exam Questions Data Analysis You think you have a fair coin. You toss it 80 times and get 47 heads. State the correct hypothesis to test. Answer: Hypothesis is p=.5 Test the hypothesis using the single sample chi-square test and a .05 a level. Answer: Chi-square = 2.45 and with 1 df you would fail to reject the hypothesis (critical value is 3.84). Test the hypothesis using the normal approximation to the binomial and a .05 a level. .5875 - .5 Answer: t = = 1.565 so you would fail to reject (critical value of t is 1.96) (.5)(.5) 80 Compare your results in B and C. Suppose you took a sample of 250 (split equally between males and females). On one attitude question the females averaged 2.3 and the males averaged 4.1 (on a five point scale). The population contains 7,700 females and 3,300 males. What is your best estimate of the population mean? (2.3)(7700) + (4.1)(3300) Answer: It would be = 2.84 11000 You are interested in usage of stock brokers. You believe (hypothesize) that one-third of a population never use a broker, one-half infrequently use a broker, and one-sixth often use a broker. You take a sample of 144 people and find that 38 never use a broker, 74 infrequently use a broker, and 32 often use a broker. Test the hypothesis at the .10 level of significance. (NOTE: This is virtually identical to my of testing the fairness of a die). Observed Expected Never 38 48 Infrequently 74 72 Often 32 24 Chi-square = 4.8 and you would reject the hypothesis (critical value with 2 df is 4.6). Actually, the interpretation is that people use brokers more than expected. You hypothesize that the mean age of a population is 32. You take a sample of 200 and find that the mean is 33 and the variance of individual ages is 60. Test the hypothesis at the .05 level of significance. Answer: t = 1.83 and you would fail to reject. How would your answer change (if at all) if you tested at the .10 level of significance? Answer: You would reject (since the critical value is 1.65) and you would be taking a 10 percent chance of rejecting a true hypothesis. What would you calculate as your 90 percent confidence interval? Answer: 33 1.65 60 200 You are developing a marketing campaign for a new product oriented toward young singles. You will introduce the product only if the proportion of young singles is at least fifteen percent of the population. What statistical hypothesis would you test? Answer: Hypothesis is p .15 A newspaper reported that 40 percent of the population liked the job Clinton was doing (and 60 percent didn't). They indicated that the results were accurate within plus or minus 2 percentage points. Their sample size was 985. What is their implied level of confidence? Answer: Well, Z You want to determine the average age of undergraduates on the Boulder Campus. You take a sample of 150 undergraduate students and find the following. What is your 90 percent Confidence Interval? Briefly explain what this means. _ X = 20.5 Answer: 20.5 1.65 ^s 2 = 20 20 or 19.9 to 21.1 150 You took a sample of 500 and found that 73 percent of them felt that members of Congress should have term limits. What is your 90 percent confidence interval on this estimate? Answer: .73 + 1.65 (60)(40) so Z = 1.28 which is 80 percent confidence. = 2 985 (.73)(.27) 500 or .6872 to .7628 I gave you a four step procedure for hypothesis testing. Describe this procedure. Give a simple example which clearly demonstrates this procedure. You take a sample of a population and find the following. ATTITUDE Low 23 22 Male Female Medium 37 54 High 44 35 State the hypothesis you would test. Answer: There is no relationship between gender and attitude. Test the hypothesis at the .05 level of significance. Answer: The expected values are Male Female Low 21.8 23.2 45 Medium 44.0 47.0 91 High 38.2 40.8 79 104 111 215 The Chi-square value is 3.9895 and, with a critical value of 5.99, you would fail to reject the hypothesis. How would your answer change (if at all) if you tested at the .10 level of significance? Answer: At the 10 percent level, the critical value is 4.605 and you would still fail to reject. You basically would not include the table in a report, because there is no significant relationship. Assume you chose to include this information in a report. Show the table you would include in the report, and state how you would describe your interpretation of this table. ATTITUDE Male Low 22 Medium 36 High 42 Female 20 49 31 Chi-Square = 3.99 100% (n = 104) 100% (n = 111) Significance = 15 percent While one-third of the males have a medium attitude, nearly one-half of the females have a medium attitude. More males (43 percent) have a high attitude than do females (31 percent). You have a coin which you believe is loaded so that heads come up twice as often as tails. You toss the coin 70 times and get 38 heads. State the correct hypothesis to test. Answer: Hypothesis is p = .67 Test the hypothesis using the single sample chi-square test and a .05 a level. 2 You would expect 463 heads. The Chi-square would be 4.83 and you would reject the hypothesis (critical value is 3.84) Test the hypothesis using the normal approximation to the binomial and a .05 a level. .543 - .666 Answer: t = = -2.19 so you would reject the hypothesis. (.67)(.33) 70 Compare your results in B and C. A friend told you he had tested a 2 by 3 cross tabulation table and got a chi-square value of around 3.8 and he thought a significance value of around .02 or so. What would you tell him? Answer: Not possible . . . he screwed something up. There are 2 df, and the critical value at .02 is about 7. The value of 3.8 would be significant around .15 or so. A study of Cola drinking yielded the following table. Male Female Cola Liked Best Pepsi Classic Coke 23 7 16 9 New Coke 10 17 State the hypothesis which you would test. State the alternative. Answer: The hypothesis is there is no relationship between gender and choice of cola. The alternative is that there is a relationship. What is the calculated Chi-Square value. At what risk (alpha) level would you reject the hypothesis. Answer: The Chi-square value is 3.32 so you would reject at around the .17 level of significance. Assume there is a significant relationship. Interpret this table. A manufacturer of digital audio tape players surveyed 100 retail stores in each of his sales regions. An analyst noted that in the Southern region the average retail price was $165 and the standard deviation was $30, while in the Atlantic region the average price was $170 with a standard deviation of $15. What do these statistics tell us about these two regions? Assume you have the following data: H0: = 200 n = 64 s = 30 sample mean = 218 Conduct a two-tailed hypothesis test at the .05 significance level. 218 - 200 Answer: t = 30 = 4.8 and you would reject the hypothesis. 64 Calculate a 90 percent confidence interval. 302 Answer: 218 1.64 64 or 211.85 to 224.15. Note that the number given in the problem (30) is standard deviation, not variance. You hypothesize that the average age of undergraduates on the Denver Campus is 30½ years, and want to test this hypothesis at the 10 percent level of significance. You take a sample of 150 undergraduate students and find the following? Would you reject or fail to reject? Why? _ ^s 2 = 49 X = 29.5 29.5 - 30.5 Answer: t = = 1.75 and you would reject the hypothesis. 49 150 You are measuring the value of new automobiles sold in Boulder (in thousands of dollars) and want to calculate a 90 percent confidence interval. You took a sample of 75 people who recently purchased an automobile and came up with the following results. What is your confidence interval? _ ^s 2 = 2.5 X = 17.5 Answer: 17.5 1.64 2.5 75 Briefly describe alpha and beta errors. You are interested in whether males and females have the same attitudes, and want to test this at the 10 percent level of significance. You took a sample of 150 males and 200 females and got the following results. Would you reject or fail to reject? Why? _ Xmales = 5.5 _ Xfemales = 6.0 ^s 2males = 9 ^s 2females = 16 Answer: I’ll work this some other day since we did not discuss this test. You have done a survey of purchasing behavior and created the following table of counts relating purchasing and gender. You are going to do a Chi-square test at a 10 percent level of significance. Males Females Almost Never Purchase 67 45 Purchase Occasionally Frequently Purchase 23 67 44 55 State the hypothesis you are going to test. Answer: There is no relationship between gender and purchase behavior. Would you reject or fail to reject? Why? Answer: Chi-square is 23.72 so you would reject big time. Actually, only needed to calculate the first cell (which gives a value over 5) to see that you would reject. The reason for rejection is that 23.72 is highly unexpected and much bigger than the critical value. (APPLIES TO ALL FOUR QUESTIONS) You are interested in taking a sample of Boulder to determine the average income. You expect it to be around $38,000. However, knowing that the distribution of income is skewed to the right, you expect half of the people to have income above $30,000 and half below. You also expect that virtually everyone will have an income between $10,000 and $100,000. 1. How big of a sample should you take to be 90% sure your estimate of the mean income is correct within $1,500? Answer: You expect the standard deviation to be 15,000 (the range divided by 6). Then 2 2 n = Z = 273 E2 (APPLIES TO NEXT TWO QUESTIONS) Suppose you took a sample of size 250 and found that the mean income was 35,000 and the variance of income was $25,000,000. 2. What is your 95% confidence interval on the estimate of the mean income? Answer: 35,000 1.96 25000000 250 or 34,380 to 35,620 3. Based on these sample results, you are asked to place a bet on the income of a single individual chosen at random from the population. Within what range is there a 50% chance that this individual's income would fall? Answer: This is a trick question. It deals with the range for an individual chosen from the population. The population standard deviation is 5,000 and 50 percent of all observations fall within .675 standard deviations, so you would bet that any individual’s income has a 50 percent chance of falling between 31,625 and 38,375. (APPLIES TO FINAL QUESTION) You are asked to include a question concerning the presence of children in the household. You expect that 40% of the households have children living at home (and 60% don't). 4. How big of a sample would you have to take to be 99% sure your estimate of the proportion of households having children living at home is correct within 5 percentage points? 2.582 (.6)(.4) Answer: n = or 639. This is one of those places where people will 2 .05 take (.6)(.4) and then try to square it again, but that is the variance. The other mistake made is to use 5 in the denominator rather than .05 (you have to keep the terms consistent . . . you can use 5 in the denominator but then must use 60 and 40 in the numerator). I always set up problems to give reasonable answers. If the answer just can’t be right, it probably isn’t. Basic formula for the Chi-square test Basic formula for confidence intervals (Observed Expected )2 Expected xZ Sample size adjustment for finite population sˆ 2 n n' = (N * n) (N + n - 1) Formulas for estimating population variance Normal Uniform b-a 6 2 (b - a) s = 12 s= Proportion 2 2 s = p*q Points on a Normal Distribution One Tail Two Tail 0 0.68 0.84 1.28 1.65 2.33 0.68 1.15 1.28 1.65 1.96 2.58 50 % 25 % 20 % 10 % 5% 1% Points on the Chi-square Distribution df 1 2 3 4 5 6 7 8 9 0.2 1.642 3.219 4.642 5.989 7.289 8.558 9.803 11.030 12.242 0.15 2.072 3.794 5.317 6.745 8.115 9.446 10.748 12.027 13.288 0.1 2.706 4.605 6.251 7.779 9.236 10.645 12.017 13.362 14.684 0.05 3.841 5.991 7.815 9.488 11.070 12.592 14.067 15.507 16.919 0.025 5.024 7.378 9.348 11.143 12.832 14.449 16.013 17.535 19.023 0.01 6.635 9.210 11.345 13.277 15.086 16.812 18.475 20.090 21.666 Revised: September 28, 1999 / [email protected] / University of Colorado - Boulder 0.005 7.879 10.597 12.838 14.860 16.750 18.548 20.278 21.955 23.589