4. (a) The p-value is 0.2288.
(b) Since the p-value (0.2288) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Therefore, we cannot conclude that the relationship is signficant.
(c) Since the results are not significant, we cannot use the regression line to make predictions.
5. y = 25.99
Age | Deaths |
17.5 | 38 |
22 | 36 |
29.5 | 24 |
44.5 | 20 |
64.5 | 18 |
80 | 28 |
r² | 0.335 | |||||
r | -0.579 | |||||
Std. Error | 7.533 | |||||
n | 6 | |||||
k | 1 | |||||
Dep. Var. | Deaths | |||||
ANOVA table | ||||||
Source | SS | df | MS | F | p-value | |
Regression | 114.3276 | 1 | 114.3276 | 2.01 | .2288 | |
Residual | 227.0057 | 4 | 56.7514 | |||
Total | 341.3333 | 5 | ||||
Regression output | confidence interval | |||||
variables | coefficients | std. error | t (df=4) | p-value | 95% lower | 95% upper |
Intercept | 35.5818 | |||||
Age | -0.1918 | 0.1352 | -1.419 | .2288 | -0.5671 | 0.1834 |
4. Can we use the regression line to make reliable prediction of the number of deaths...
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In determining if this regression is significant, I observed the
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To check if your results are reliable (statistically
significant), look at Significance F (0.00). If this value is less
than 0.05, the regression is acceptable. If Significance F is
greater than 0.05, it's advisable to stop using this set of
independent variables.
As part of the hypothesis test, we should evaluate R-squared as
it measures the strength of the relationship between the model...