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# hw11_solution - 17.19(refer to 17.5 Let x1 be the age of...

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17.19 (refer to 17.5) Let x 1 be the age of employee. Let x 2 be the number of years employee has stayed. Let x 3 be the annual pay (in thousands of dollars) of employee. Let y be the number of weeks of severance pay. According to the given data, we find the sample regression line as: y = -0.0078 x 1 + 0.6035 x 2 – 0.0703 x 3 a. H 0 : errors are normally distributed H 1 : errors are not normally distributed Since the number of standard residuals is larger than 30, we use chi-squared test for normality. With alpha = 0.05, we reject null hypothesis if statistic p-value < 0.05. All expected values are larger than 5. The “rule of five” is satisfied and it is valid to use the chi-squared test. Since p-value = 0.1225 > 0.05 , we fail to reject the null hypothesis. Thus, we conclude with 95 % confidence that there is insufficient evidence to infer errors are non-normal. b. Scatttterr pploott oof pprredictted yy vverrssuuss ssttanndarrd rressiduualss -4 -3 -2 -1 0 1 2 3 0 2 4 6 8 10 12 14 16 18 Prredictted yy Sttanndarrd rressiduual The spread of points doesn’t change much when predicted y varies. There is little sign of heteroscedasticity.

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c. The correlation between age and years is high (more than 0.7) indicating that multicollinearity could be a problem. 17.27 (refer to 17.13) Define: y = amount spent on lottery tickets as a percentage of total household income x 1 = number of years of education x 2 = age x 3 = number of children x 4 = personal income (in thousands of dollars) The sample regression line: y = 11.9061 - 0.4300 x 1 + 0.0292 x 2 + 0.0934 x 3 - 0.0745 x 4 a. H 0 : errors are normally distributed H 1 : errors are not normally distributed Since the number of standard residuals is larger than 30, we use chi-squared test for normality. With alpha = 0.05, we reject null hypothesis if statistic p-value < 0.05. All expected values are all larger than 5, so the “rule of five” is satisfied. It’s valid to use chi-squared test. Since p-value = 0.0154 < 0.05 , we reject the null hypothesis. Thus, we conclude with 95 % confidence that there is sufficient evidence to infer errors are non-normal. --
The spread of points get wider slightly when predicted y grows. There is minor sign of heteroscedasticity. -- H 0 : randomness exists among errors H 1 : randomness does not exist among errors With alpha = 0.05, we reject null hypothesis if statistic p-value < 0.05. Since two-tailed p-value = 1 > 0.05 , we fail to reject the null hypothesis. Thus, we conclude with 95% confidence that there is insufficient evidence to infer randomness doesn’t exist among errors. b. There is a strong correlation between income and education (the coefficient of correlation > 0.7). The t–tests of these two coefficients may be distorted.

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17.35 a. The sample regression line: y = 303.2529 + 14.9430 x 1 + 10.5211 x 2 b. Scattttooerr pploott oof ssttanndarrd rressiduualss vverrssuuss tthe ttimme pperrioodss -3 -2 -1 0 1 2 3 0 20 40 60 80 100 Timme pperriood Sttanndarrd rressiduual The graph indicates that autocorrelation exists c.
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