Question

A physical education department wanted a single test of upper body strength that was easy to administer. Dips on the parallel bars and pull-ups on the horizontal bar were considered good tests. One faculty member thought that both tests were not needed because the correlation between the two was probably high. To evaluate this assumption, 141 students were tested on both criteria. The faculty member let X represent dips on the parallel bars and Y represent pull-ups and calculated the following from the data:

ΣX = 3,416, ΣY = 1,899, SDX = 8.84, SDY = 4.70, ΣZXZY = 109.416.

Calculate the following:

a. The mean of each variable

b. The correlation between the two variables

c. The level of confidence and the p value reached by the coefficient

d. The predicted number of pull-ups for a student who performed 20 dips

e. The standard error of the estimate

f. The predicted range of possible pull-up scores at the 95% LOC for a student who performed 20 dips

Answer 1

a) $\small \bar{x}=24.227,\ \bar{y}=13.468$

b) The correlation between the two variables is

$\small \sum \frac{Z_x*Z_y}{140}=0.781543$

c) Let the level of significance be $\small \alpha=$ 0.05. Then the test statistic is given by:

$\small T=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}=14.77$. The test statistic follows a t-distribution with n-2 = 139 degrees of freedom. The p-value here is approximately 0. Thus we conclude that there is a significant positive linear relationship between the two variables.

d) First, we will find the least square regression line. The slope of the lines is:

$\small slope=\frac{\sum Z_x*Z_y}{SD_x^2}=1.4$ while the y-intercept is given by:

$\small y-intercept=\bar{y}-slope*\bar{x}=-20.4534$. Thus the regression line is pull-ups = -20.4534+1.4*dips. Thus for a student who performed 20 dips the estimated number of pull-ups is 7.5466.