| Tag | Value |
|---|---|
| file | Inferential_Statistics_vufsw-confidenceinterval-1240-en_vufsw-confidenceinterval-1240-en |
| name | vufsw-confidenceinterval-1240-en |
| section | inferential statistics/regression/confidence interval |
| type | schoice |
| solution | TRUE, FALSE, FALSE, FALSE |
| Type | calculation |
| Program | calculator |
| Language | English |
| Level | statistical literacy |
Ten students have been randomly selected. It is checked whether the number of times per week that they go out (X) is related to their physical health (Y). The regression results are below. Use the results to find the 95% confidence interval around the slope that is found in the sample.
The regression formula is Y = 3.58 + 0.09x with 8 degrees of freedom
| Predictor | b (Coef) | SEb | t | p |
| Constant | 3.58 | .09 | 45.85 | .00 |
| go out | .09 | .02 | 4.50 | .00 |
You can calculate this with the following formula: CI for B
+/- t(se)
The t value is 2.306 (see table: t Distribution Critical Values).
Thus: 0.09 +/- 2.306 * 0.02 = (0.04; 0.14)
M&T Bivariate linear regression
Bivariate linear regression
M&T Multivariate linear regression
Default value