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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

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Answer:

24.23 ; 13.47 ; 0.782 ; p value < 0.00001 ;7.45

Step-by-step explanation:

Given that :

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

Sample size, N = 141

Mean of X:

£X / N = 3416 / 141 = 24.23

Mean of Y:

£Y / N = 1899 / 141 = 13.47

B.) Correlation between the 2 variables :

£ZXZY / N - 1

= 109.416 / 140

= 0.782

C.)

Level of significance and p value :

T = [Rsqrt(n-2)] ÷ sqrt(1 - R²)

T = [0.782 * sqrt(139)] ÷ sqrt(1-0.782^2)

T = 14.79

P value from Tscore calculator

Df = 141 - 2 = 139 ; α = 0.05

Pvalue = 0.00001

Pvalue < α ; Hence, a significant positive relationship exists.

To obtain regression model :

Slope = £ZXZY / SDX²

m = 109.416 / 8.84^2 = 1. 4

yintercept = meanY - slope*x

yintercept = 13.47 - 1.4*24.3

yintercept = - 20.55

Equation in slope intercept form:

y = mx + c

y = 1.4x - 20.55

For x = 20

y = 1.4(20) - 20.55

y = 7.45

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