Business statistics 2, Exercises of Business Statistics

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Answer to question number 1:

Step 1: State null and alternative hypothesis

The null hypothesis (H 0 ) = Professional couples live in homes of the same size as their parents. The alternative hypothesis (H 1 ) = Professional couples live in larger homes than their parents.

Step 2: Select level of significance

The level of significance is α = 0.05 for a one-tailed test.

Step 3: Identify test statistics

The Wilcoxon signed-rank test is used. The differences between the paired observations are calculated, absolute values are ranked, and signs are returned to the ranks. The test statistic T is the smaller of the two sums of signed ranks.

Step 4: Statistical test and formulate decision rule

Table of differences and ranks: Couple Name Professional Parents Difference (D) Absolute Difference Rank Rank(+) Rank(-) Gordon 1,725 1,175 +550 550 7 7 Sharkey 1,310 1,120 +190 190 5 5 Uselding 1,670 1,420 +250 250 6 6 Bell 1,520 1,640 - 120 120 3 3 Kuhlman 1,290 1,360 - 70 70 1 1 Welch 1,880 1,750 +130 130 4 4 Anderson 1,530 1,440 +90 90 2 2 Sum of positive ranks = 24 Sum of negative ranks = 4 Test statistic T = smaller of the two sums = 4 Decision rule: For n = 7 and α = 0.05 (one-tailed), the critical value is Tcritical = 3. Reject H 0 if T ≤ 3.

Step 5: Conclusion

Since T = 4 > 3, we do not reject the null hypothesis. Therefore, at the 0.05 significance level, there is no evidence to conclude that professional couples live in larger homes than their parents.

Step 5: Conclusion

At the 0.01 significance level, we do not find evidence of a difference in the lasting quality of the epoxy paint among the three water conditions. Therefore, the data do not indicate a statistically significant difference across saltwater, freshwater without weeds, and freshwater with a heavy concentration of weeds.

Answer to question number 3 : The logarithmic trend equation for the turnovers data: log Ȳ = log a + log b(t) log Ȳ = 1.00455 + 0.04409(t) Here, log b = 0.04409 and its antilog is 1.106853. The value (1.106853 − 1) = 0.106853 indicates the geometric-mean annual rate of increase in asset turnovers from 2012 to 2022. This means turnovers increased by 10.685% annually during this period. To estimate the projected turnover for the year 2024, we use the coded value t = 13. Substituting into the equation: log Ȳ = 1.00455 + 0.04409(13) log Ȳ = 1. Taking the antilog of 1.57772 gives: Ȳ = 10^1.57772 = 37. So, the projected turnover amount for 2024 is 37.8. The data table Year Asset turnover Code 2012 1.11 1 2013 1.28 2 2014 1.17 3 2015 1.10 4 2016 1.06 5 2017 1.14 6 2018 1.24 7 2019 1.33 8 2020 1.38 9 2021 1.50 10 2022 1.65 11 2024 13

Answer to question number 5 : a) The regression equation is Ŷ = 84.998 + 2.391X₁ − 0.4086X₂. b) For X₁ = 4 and X₂ = 11, the estimated value of the dependent variable is: Ŷ = 84.998 + 2.391(4) − 0.4086(11) = 84.998 + 9.564 − 4.4946 = 90.0954. c) The total degrees of freedom reported in the ANOVA table are 64, so the sample size is n = 64

• 1 = 65. There are two independent variables, X₁ and X₂. d) Step 1: State null and alternative hypothesis H 0 : β 1 = β 2 = 0 (the model has no explanatory power). H 1 : Not all β’s is equal to 0 (at least one coefficient differs from zero). Step 2: Select level of significance The significance level is α = 0.05. Step 3: Identify test statistics Use the overall F-test from the ANOVA. The reported F statistic is Fcal = 4.14 with degrees of freedom (2, 64). Step 4: Statistical test and formulate decision rule The critical value at α = 0.05 for df = (2, 64) is Fcritical = 3.1 5. The decision rule is: reject H 0 if Fcal

Fcritical. Since 4.14 > 3.1 5 , we reject H 0. Step 5: Conclusion We reject the null hypothesis and conclude that not all regression coefficients are zero; the overall regression model is statistically significant at the 0.05 level.

e) Step 1: State null and alternative hypothesis For X 1 : H 0 : β 1 = 0 versus H 1 : β 1 ≠ 0. For X 2 : H 0 : β 2 = 0 versus H 1 : β 2 ≠ 0. Step 2: Select level of significance Use α = 0.05 for both tests. Step 3: Identify test statistics Rely on the reported p-values from the regression output: p(X 1 ) = 0.051 and p(X 2 ) = 0.020. Step 4: Statistical test and formulate decision rule Decision rule: reject H 0 if p-value < 0.05. For X 1 : 0.051 > 0.05, so we fail to reject H 0 ; X 1 is not statistically significant at the 0.05 level and is a candidate for removal. However, because p < 0.10, there is weak evidence of a relationship, and retaining X 1 can be considered depending on model objectives. For X 2 : 0.020 < 0.05, so we reject H 0 ; X 2 is statistically significant. Step 5: Conclusion At α = 0.05, X 1 is not significant (p = 0.051) and could be considered for elimination, while X 2 is significant (p = 0.020) and should be retained. f) A reasonable variable selection strategy is to drop the variable with the highest p-value that is not statistically significant at the chosen α level. Since X₁ has p = 0.051 and is not significant at 0.05, it should be considered for elimination first.

Step 5: Conclusion

Since 115.224242 > 11.345, reject H 0. At the 0.01 significance level, the chi-square test indicates a significant difference between the observed and expected distributions of education levels among cardholders who failed to pay. Therefore, the distribution of nonpaying cardholders is different from the historical distribution.