hw09_solution - 16.69 The independent variable (x) is the...

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16.69 The independent variable (x) is “the floor number of the condominium” and the dependent variable (y) is “the price (in $1,000)”. According to the given data, we find the regression line: 3734 . 190 4654 . 1 ˆ + = x y . 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 = 0.5079 > 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. Scatter diagram of standard residuals -3 -2 -1 0 1 2 3 150 170 190 210 230 250 Predicted dependent variable (Price) Standardresiduals The spread of points doesn’t change much when price varies. There is little sign of heteroscedasticity.
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16.77 The independent variable is “annual income (in $1,000)” and the dependent variable is “weekly spent on food”. According to the given data, we find the corresponding regression line: 8986 . 153 9582 . 1 ˆ + = x y . (i) First, we check whether errors are normally distributed: 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 = 0.3618 > 0.05 , we fail to reject the null hypothesis. Thus, we conclude with 95% confidence that is insufficient evidence to infer errors are non-normal. (ii) Next, we draw scatter diagram to diagnose heteroscedasticity:
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hw09_solution - 16.69 The independent variable (x) is the...

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