MHB F-Test : Do we accept the null-hypothesis?

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The discussion revolves around performing an F-test to determine if the slope of a regression line, which predicts monthly gross income based on years of service, is significantly different from zero. The calculated F-value is 0.0997, leading to a p-value of approximately 0.754, indicating insufficient evidence to reject the null hypothesis. Participants clarify that while the null hypothesis is retained, it does not confirm that the slope is exactly zero. Suggestions for verifying results include using Excel's LINEST function and conducting a t-test for the slope, which should yield consistent p-values. The conversation emphasizes the importance of checking calculations and understanding the implications of statistical results.
mathmari
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Hey! :o

The below table shows the average monthly gross income of a sample of 44 developers. For each individual sample, it is indicated their country of employment and years of service in their field.

Calculate the regression line with the dependent variable the monthly gross income and independent the years of employee service and check the significance of the F criterion at $\alpha = 0.05$. I have done the following:

View attachment 9503

Therefore we get:
\begin{align*}\nu &=44 \\ \overline{X}&=\frac{\sum X}{\nu}=\frac{426.1}{44}=9.68 \\ \overline{Y}&=\frac{\sum Y}{\nu}=\frac{299767.60}{44}=6812.9 \\ \hat{\beta}&=\frac{\nu \sum \left (XY\right )-\left (\sum X\right )\left (\sum Y\right )}{\nu\sum X^2-\left (\sum X\right )^2}=\frac{44 \cdot 2911490.795-426.1\cdot 299767.60}{44\cdot 4417.39-426.1^2}=\frac{1.2810559498 \cdot 10^8-1.2773097436 \cdot 10^8}{194365.16-181561.21} \\ & =\frac{374620.62}{12803.95}=29.26 \\ \hat{\alpha}&=\overline{Y}-\hat{\beta}\cdot \overline{X}=6812.9-29.26\cdot 9.68=6812.9-283.2368=6529.66\end{align*}

So the regression line is \begin{equation*}\hat{Y}=29.26X+6529.66\end{equation*} We consider a F-test whether the slope is $0$ or not.

We have the formula $\displaystyle{F=\frac{MSM}{MSE}=\frac{\text{explained variance}}{\text{unexplained variance}}}$.

We have that $\displaystyle{MSM = \frac{SSM}{DFM}}$ with $\displaystyle{SSM=\sum_{i=1}^n\left (\hat{Y}_i-\overline{Y}\right )^2}$ and $\displaystyle{DFM = p - 1}$.

We also have that $\displaystyle{MSE = \frac{SSE}{DFE}}$ with $\displaystyle{SSE=\sum_{i=1}^n\left (Y_i-\hat{Y}_i\right )^2}$ and $\displaystyle{DFE = \nu-p}$. We have this table.

So we get
\begin{align*}&SSM=249138.5759 \\ &DFM=2-1=1 \\ &SSE=104926827.5 \\ &DFE=44-2=42 \\ &MSM=\frac{SSM}{DFM}=\frac{249138.5759}{1}=249138.5759 \\ &MSE=\frac{SSE}{DFE}=\frac{104926827.5}{42}=2498257.7976 \\ &F=\frac{MSM}{MSE}=\frac{249138.5759}{2498257.7976}=0.0997\end{align*}

Using at the R-program the command pf(0.0997, 1, 42, lower.tail=F) we get the p-value $0.7537537$.

That means that $\text{p-value} > \alpha$, does this mean that we accept the null hypothesis, i.e. that the slope is not significally different from $0$.

Is that correct? (Wondering)
 

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mathmari said:
That means that $\text{p-value} > \alpha$, does this mean that we accept the null hypothesis, i.e. that the slope is not significally different from $0$.

Is that correct?

Hey mathmari!

Formally it means we do not have sufficient evidence to conclude the slope is different from 0.
We 'keep' the null hypothesis, but we cannot conclude that the slope is 0. (Nerd)
 
Klaas van Aarsen said:
Formally it means we do not have sufficient evidence to conclude the slope is different from 0.
We 'keep' the null hypothesis, but we cannot conclude that the slope is 0. (Nerd)

Ah ok! Is there a way to check if my results are correct? (Wondering)
 
mathmari said:
Ah ok! Is there a way to check if my results are correct?

Excel has the [M]LINEST[/M] function.
One of its outputs is the F-value when used in a 5x2 array context for a simple linear regression. (Thinking)
Excel also has the [M]F.DIST[/M] function to convert an F-value into a probability.

We can also do a t-test for the slope as we did before. It should yield the same p-value. (Thinking)

Additionally we can draw a graph of the points and the line to see if it makes sense that the explained variance is so much lower than the unexplained variance. (Thinking)
 
Klaas van Aarsen said:
Excel has the [M]LINEST[/M] function.
One of its outputs is the F-value when used in a 5x2 array context for a simple linear regression. (Thinking)

Using the command [M]LINEST(D2:D45;B2:B45)[/M] I get $29.25833947$. If I used the command correctly, I must have a mistake at the calculations of $F$. But what? (Wondering)
 
mathmari said:
Using the command [M]LINEST(D2:D45;B2:B45)[/M] I get $29.25833947$. If I used the command correctly, I must have a mistake at the calculations of $F$. But what? (Wondering)

To get the F-value from LINEST, we need to specify 4 parameters. And we must make it an array formula with Ctrl+Shift+Enter. (Thinking)

You should also get for instance the slope and the y intercept. Do they match? (Wondering)
 
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