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C.I..95=±t(SEM)+MThe formula for the confidence interval of the population mean. Formula 11.2C.I..95=±t(SEd)+(M1?M2)The formula for the confidence interval of the difference.Formula 11.3C.I..68=±1(SEest)+y?The formula for the .68 confidence interval of the prediction.Formula 11.4C.I.=±tn?2(SEest)+y? The formula for the confidence interval of the prediction.Unless otherwise stated, use p = .05 in all your answers.1. A hardware retailer has averaged sales of $64,235, with a standard deviation of $5,918, for a 12-month period. The mean monthly sales for all retailers in the chain are $59,844. Are this hardware retailer’s sales significantly different from those of all retailers in the chain at p = .05? Are they significantly different at p = .01?2. Calculate a .95 confidence interval for the data in problem 1. Explain your findings in lay terms. What question does the confidence interval answer?3. Calculate C.I..99 for the data in problem 1. What is the effect of changing the confidence level on the width of the interval? Why?4. There are two groups of eight sales representatives for a pharmaceutical company in two regions of the country. For group 1, the average monthly charges to expense accounts are $387, with s = $22.73. For group 2, the average is $344, with s = 18.19. Is the difference between the two groups statistically significant at p = .05? Are they significantly different at p = .01?5. Calculate a C.I..95 for the difference for the item 4 data. Which value(s) represent the point estimate of the difference, and which value(s) indicate(s) the interval estimate of the difference for the item 4 data?6. What could be done to narrow the interval estimate for the item 4 data?7. A car dealer is using the number of years customers have owned their vehicles to predict how long it will be before they elect to replace them. The correlation between the two is rxy = ?.723 (the longer they have owned their present vehicles, the more quickly they are expected to replace them). The other relevant data are as follows for 32 customers: MeanStandard DeviationYears owned present vehicle4.5511.627Time until expected replacement2.2591.3389. How long will it be until a customer who has owned the vehicle 6.5 years is likely to replace it?10. Calculate .95 and .99 confidence intervals for the prediction in item 7.11. How will a larger standard deviation in the criterion variable affect the width of the confidence intervals in item 8? Why?12. What does a .95 confidence interval of the prediction for the item 7 data reveal? Chapter FormulasFormula 12.1?2=?(fo?fe)2feis the formula for the chi-square test statistic. The same formula is used for both “goodness of fit” test and for the r × k chi-square test of independence. Formula 12.2?=(X2/n)Phi coefficient is the measure of the correlation for two nominal variables when the chi-square test of independence indicates a significant result, and when one of the variables involved has two or fewer categories. Formula 12.3V?2/(smaller of rows or columns)?1When the test of independence is significant and both variables have at least three categories, Cramer’s V is calculated rather than phi coefficient. V requires phi, however, which must be calculated first.Management Application ExercisesUnless otherwise stated, use p = .05 in all your answers.Fantasy Haven: 22Night of Terror: 18Fists of Glory: 12verbal reinforcement seminar—17 questions/commentstangible reward seminar—27 questions/comments Link to textbookhttp://www.homeworkmarket.com/sites/default/files/qx/14/12/13/11/bus308_chapter_12.pdfPurchase the answer to view it
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