Question

please find the degrees of freedom,critical values, correlation coefficient, and test statistic as well.

Score: 0 of 1 pt 8 of 8 (6 complete) HW Score: 67.86%, 8.14 of 12 pts 9.1.35-T Assigned Media Question Help The maximum weights (in kilograms for which one repetition of a half-squat can be performed and the jump heights (in centimeters) for 12 international soccer players are given in the accompanying table. The correlation coefficient, rounded to three decimal places, is r=0.715. At a = 0,05, is there oncugh evidence to conclude that there is a significant linear correlation between the variables? Click the icon to view the soccer player data Determine the null and alternative hypotheses Hoplo Holo Click to select your answers) and then click Check Answer parts remaining Clear All Check Answer • Previous Next

Answer 1

Ans : Given data are

Maximum weight (x)	Jump Height (y)
190	60
185	57
155	55
180	59
175	55
170	65
150	52
160	50
160	48
180	57
190	60
210	64

Now to calculate the Correlation coefficient :

Σ(x – Mx)(y – My) r = Σ(x – Mx)? * Σy – My)?

Maximum weight (x)	Jump Height (y)	(x-Mx)*(y-My)	(x-Mx)^2	(y-My)^2
190	60	46.18056	212.6736	10.02778
185	57	1.597222	91.84028	0.027778
155	55	37.43056	416.8403	3.361111
180	59	9.930556	21.00694	4.694444
175	55	0.763889	0.173611	3.361111
170	65	-44.2361	29.34028	66.69444
150	52	122.8472	646.0069	23.36111
160	50	105.3472	237.6736	46.69444
160	48	136.1806	237.6736	78.02778
180	57	0.763889	21.00694	0.027778
190	60	46.18056	212.6736	10.02778
210	64	247.8472	1196.007	51.36111
		710.8333	3322.917	297.6667

Mx = 175.4167 My = 56.8333

710.8333 r = 3322.917 * 297.6667

r = 0.714732

the Correlation coefficient : r = 0.715 (rounded three decimal places )

Given claim that the there is a significant linear correlation between the variable

Test of hypothesis :

H₀ : $\small \rho$ = 0 versus H_a : $\small \rho \neq$ 0

Test Statistic :

r *vn - 2 t= V1 - p2

t= 0.715 * 12 - 2 V1 -0.7152

t = 3.23

Test Statistic :

t = 3.23

Critical region : If | t | > t_{n-2,$\small \alpha$/2}

here given values of $\small \alpha$ = 0.05

degrees of freedom = n-2 = 12-2 = 2

degrees of freedom = 10

Critical value = t_10,0.05  = 2.228

Critical value = 2.228

Since |t | = 3.23 > t_10,0.05  = 2.228 Then we reject H0 at 0.05 significance level

conclusion : here we reject the null hypothesis , there is not sufficient evidence to there is a significant linear correlation between the variable at 0.05 significance level .