Here is the summary of my linear model:
Call:
lm(formula = weight ~ height, data = height and weight)
Residuals:
Min 1Q Median 3Q Max
-10.267 -4.267 -1.267 6.455 11.538
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 20.4878 5.8335 3.512 0.00158 **
height 0.3195 0.1447 2.208 0.03596 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 6.703 on 27 degrees of freedom
Multiple R-squared: 0.1529, Adjusted R-squared: 0.1215
F-statistic: 4.873 on 1 and 27 DF, p-value: 0.03596
And here is my confidence interval for $\hat{\beta}_1$: 0.02260012 0.61639988 at $5\%$ level.
As you can see $0 \notin $CI at $5\%$ level. Also, the $p$-value for the test $H_0: \beta_1=0$ turns out to be $0.03596 $ which is in between $0.01 <0.03596 < 0.05$.
How do I interpret this?
The p-value < 0.05 should mean that there is moderate evidence against the null hypothesis $H_0: \beta_1=0$, also, $0$ is not in CI. So should I reject the fact that $\beta_1=0$ or should I say the evidence is not strong enough?
Furthermore, what would happen if in the same situation, I would get $0 \in CI$ but still $p-$value <0.05?
What about $p-$value >0.05 but $0\in CI$?
height)? The values you quote are inconsistent with what's shown in the output you have reproduced. – whuber Apr 07 '17 at 15:22lmfunction to create a linear model. Then I usedsummaryto generate this summary, that's it. Why do you say it is inconsistent? I might have made some typos – Euler_Salter Apr 07 '17 at 15:24qt(0.975,27)you get2.051831. The I simply calculated the CI as follows $$\hat{\beta}1 \pm c\cdot se(\hat{\beta}_1)$$ where $c = t{n-2, 1-\frac{\alpha}{2}}=t_{27, 0.975} = 2.051831$ and $se(\hat{\beta}_1) = 0.1447 $ (where the standard error you get it from the summary again). Hence if you do this calculation, you get $(0.3195-2.051831 \times 0.1447, 0.3195+2.051831 \times 0.1447) = (0.0226000, 0.6163999)$ which is the CI I gave above – Euler_Salter Apr 07 '17 at 16:00