Question

5 peimts- To examine the relationship between the amount of questions on my statistics midterms and the time it takes to complete it, I gave 5 midterms # Questions Time (min 6 70 8 81 10 85 8 76 78 Hint: Σχ¡= 41, yi-390, У-345, Σ -30546, yiyi-3228 (a) State the regression equation. (p value can be within an interval) (b) At the 0.05 significance level, is the number of questions on an exam a significant predictor of the amount of time it takes to finish? Make sure to define the hypotheses. (c) IfI want an exam to last 90 minutes, how many questions should it be?

Answer 1

Let Y be the time (in mins) taken complete the midterm and X be the number of questions in the mid term.

The regression equation that we want to estimate is

$Y=\beta_0+\beta_1X+\epsilon$

where $\beta_0$ is the intercept of the regression line

$\beta_1$ is the slope of regression line

$\epsilon \stackrel{iid}\sim \mathcal{N}(0,\sigma^2)$ is a random error

We calculate the following

n=5

$\bar{x}=\frac{\sum x_i}{n}=\frac{41}{5}=8.2$

$\bar{y}=\frac{\sum y_i}{n}=\frac{390}{5}=78$

$SS_x=\sum (x_i-\bar{x})^2=\sum x^2-n\bar{x}^2=345-5\times 8.2^2=8.8$

$SS_y=\sum (y_i-\bar{y})^2=\sum y^2-n\bar{y}^2=30546-5\times 78^2=126$

$SS_{xy}=\sum (x_i-\bar{x})(y_i-\bar{y})=\sum x_iy_i-n\bar{x}\bar{y}=3228-5\times 8.2\times 78=30$

The estimate of slope is

$\hat{\beta}_1=\frac{SS_{xy}}{SS_x}=\frac{30}{8.8}=3.409$

The estimate of intercept is

$\hat{\beta}_0=\bar{y}-\hat{\beta}_1\bar{x}=78-3.409\times 8.2=50.045$

a) The estimated regression equation is

$\hat{Y}=50.045+3.409X$

b) The number of questions on an exam (X) a significant predictor of the amount of time it takes to finish (Y), if the slope coefficient corresponding to X ($\beta_1$) is not equal to zero.

We want to test the following hypotheses

$\begin{align*} &H_0:\beta_1=0\leftarrow\text{null hypothesis: the number of questions is not a predictor of amount of time taken }\\ &H_a:\beta_1\ne 0\leftarrow\text{alternative hypothesis: the number of questions is a predictor of amount of time taken }\\ &\alpha=0.05\leftarrow\text{level of significance to test the hypotheses} \end{align*}$

We need to calculate the following

Sum of square Errors is

$\begin{align*} SSE=SS_y-\hat{\beta}_1SS_xy=126-3.409\times 30=23.727 \end{align*}$

Mean square Error is

$\begin{align*} MSE=\frac{SSE}{n-2}=\frac{23.727}{5-2}=7.909 \end{align*}$

The standard error of estimate is

$\begin{align*} s=\sqrt{MSE}=\sqrt{7.909}=2.8123 \end{align*}$

The standard error of the slope estimate is

$\begin{align*} s(\hat{\beta}_1)=\frac{s}{\sqrt{SS_x}}=\frac{2.8123}{\sqrt{8.8}}=0.9480 \end{align*}$

The hypothesized value of the slope is $\begin{align*} \beta_{1H_0}=0 \end{align*}$

the test statistics is

$\begin{align*} t=\frac{\hat{\beta}_1-\beta_{1H_0}}{s(\hat{\beta}_1)}=\frac{3.409-0}{0.9480}=3.596 \end{align*}$

The degrees of freedom is n-2=5-2=3

This is a 2 tailed test (The alternative hypothesis has "not equal to"). The p value of this statistics is

P(T<-3.596) +P(T>3.596)

Using the standard t tables, we can get the p-value with in an interval.

For 3 degrees of freedom, we can see that the total area under 2 tails of t-distribution is 0.05, when t=3.182 and is 0.02 when t=4.541.

The test statistics is 3.596. That means the p-value is in the interval 0.02 to 0.5

(We can get the exact p-value in excel using =T.DIST.2T(3.596,3) as 0.037)

We will reject the null hypothesis if the p-value is less than the significance level.

Here the p-value is in the interval 0.02 to 0.05, and hence is less than the significance level of 0.05.

Hence we will reject the null hypothesis.

We conclude that at 0.05 level of significance, the number of questions on an exam a significant predictor of the amount of time it takes to finish.

c) Using the equation

$\hat{Y}=50.045&plus;3.409X$

we need to solve for X when the predicted value of Y is 90

$90=50.045&plus;3.409X\implies X=\frac{90-50.045}{3.409}=11.72$

If I want an exam to last 90 minutes, there should be 12 questions in the exam.