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Regression Statistics Multiple R 0.93154423 R Square 0.86777466 Adjusted R Square 0.83726112 Standard Error 15231.9039 Observations ANOVA df F Significance F 5.55716E-06 28.439 Regression Residual Total 3 13 16 SS MS 19794476008 6.6E+09 3016141639 2.32E+08 22810617647 Intercept Square footage Age Bedrooms Coefficients Standard Error Stat P-value Lower 95% Upper 95% 91446.493 26076.89048 3.506802 0.003863 35110.79625 147782.19 29.8578818 10.86089528 2.749118 0.016566 6.394344068 53.3214195 - 1504.7659 370.8203691 -4.05794 0.001356 -2305.874572 -703.65717 2116.85543 10003.00921 0.211622 0.835684 - 19493.3321 23727.043 From the Excel output, a regression model to predict selling price (y) on square footage, age and number of bedrooms (x's) is, j= Bo + Box: + B2x2 + B, X3 = 91446.49 +29.86x2 – 1504.77x2 +2116.86xz
The coefficient of determination (R-squared) is, From Excel output, the coefficient of determination (R-squared) is 0.8678 To test the hypothesis is that the overall regression model is significant at 5% level of significance. The null and alternative hypothesis is, Ho:B2 = B2 = B, = 0 He: At least one of B, is not significant The F-test statistics is, From the Excel output, the F-test statistics is 28.44 The p-value for this test is, From the Excel output, the p value for this test is 0.0000 Decision
The conclusion is that p-value in this context is less than 0.05 which is 0.0000, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the overall regression model is significant. The result is statistically significant. To test the hypothesis is that the slop of square footage is significantly different from 0 at 5% significance level. The null and alternative hypothesis is, H.:B. = 0 H.B.00 The t-test statistics is, From the Excel output, the t-test statistics is 2.75 The p-value for this test is, From the Excel output, the p-value for this test is 0.0166. Decision
The conclusion is that the p-value in this context is less than 0.05 which is 0.0166, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the slop of square footage is significantly different from 0. The result is statistically significant. To test the hypothesis is that the slop of age is significantly different from 0 at 5% significance level. The null and alternative hypothesis is, H :B2 = 0 HB2 0 The t-test statistics is, From the Excel output, the t-test statistics is - 4.06. The p-value for this test is, From the Excel output, the p-value for this test is 0.0014.
Decision The conclusion is that the p-value in this context is less than 0.05 which is 0.0014, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that the slop of age is significantly different from 0. The result is statistically significant To test the hypothesis is that the slop of bedrooms is significantly different from 0 at 5% significance level. The null and alternative hypothesis is, H.:B; = 0 H:8,60 The t-test statistics is, From the Excel output, the t-test statistics is 0.21]. The p-value for this test is, From the Excel output, the p-value for this test is 0.8357
Decision The conclusion is that the p-value in this context is higher than 0.05 which is 0.8357, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that the slop of bedrooms is significantly different from 0. The result is not statistically significant.