Question

MEDAL/FAN
Can a pretest on mathematics skills predict success in a statistics course? The 55 students in an introductory statistics class took a pretest at the beginning of the semester. The least-squares regression line for predicting the score y on the final exam from the pretest score x was y = 10.5 + 0.82x . The product of 0.82 times x (the pre-test score) + 10.5 predicts the score on the final exam, y. The standard error of b was 0.38. Test the null hypothesis that there is no linear relationship between the pretest score and the score on the final exam against the two-sided alternative.

Answer 1

So assuming you have the equation \(\hat{y}=\hat{\alpha}+\hat{\beta}x\), Then the hypothesis to test is to verify if the regression parameter is equal to 0 or not. I will use \(\beta\) without the "hat" for the true parameter.

\(H_0: \beta=0... \text{ "no linear relationship"}\)
\(H_a: \beta \ne 0 ... \text{" there is a linear relationship"}\)

T-statistic:
\[\large t=\frac{\hat{\beta}-\beta}{SE(\hat{\beta})}\sim t_{n-2}=t_{53} \text{ distribution}\\
\\ \, \\\large =\frac{0.82-0}{0.38} =2.1579\]

Now assuming a 0.05 significance level (it's not stated in the problem, but that's usually what is asked...) we can find the p-value using a t-table..
Or you can use a t-distribution calculator (such as here http://stattrek.com/online-calculator/t-distribution.aspx ).. also since your sample size is high, you could also probably just use a normal distribution table/calculator (which I'll do).

\[ \large \text{p-value } =P\left(|T|>2.1579 \right)\\\large =2P(T>2.1579), \text{ where } T\sim t_{53} \approx N(0,1) \]

Using this table http://www.google.ca/imgres?imgurl=&imgrefurl=http%3A%2F%2Fwww.six-sigma-material.com%2FNormal-Distribution.html&h=0&w=0&tbnid=hqtQlOirXbBiJM&zoom=1&tbnh=232&tbnw=217&docid=aRV9ahuBJoEofM&tbm=isch&ei=XXjlU_XqIIjG8AHP-YDoCg&ved=0CAIQsCUoAA

and the value 2.16, we find the p-value to be \(2(1 - 0.9846)=0.0308 < 0.05\) So since 0.0308 is less than the significance level of 0.05, we reject the null hypothesis that \(\beta = 0\) and so there is evidence to believe that the null hypothesis may be true (i.e. that there IS a linear relationship)

Answer 2

last sentence: ...that the alternative hypothesis may be true*

Answer 3

@kirbykirby YOU ARE AN ABSOLUTE LIFE SAVER THANK YOU SO MUCH <3

Answer 4

your welcome :)!