Question

Latx be a random variable that represents the betting average of a professional baseball player Lety be a random variable that represents the percentage of strikeouts of a professional baseball player. A random sample of 6 professional baseball players gave the following formation 02 0340 ta) verify thatx 1.832, Ly - 37.3, 0.570554, 22-200.91, Xx - 30.6712, and 0.391 Ey 13 Xory 10712 Use a 10% of cance to test them the 2 decimal places) critical to 148 Conclusion Reject the nut hypothesis, there is suficient evidence that differs from Reject the mul hypothesis, there is incent evidence that pers from Fail to react the upothesis, there is niet evidence that offers from o Fall to reject the pothesis, there is sufficient evidence that pers from c) very that S, - 1.7295 2 25.813, and 64179. S. 25 ( Find the predicted percentage of out for player with x 0.312 being average (Use 2 decimal places) 450 (4) Find a 10% confidence interval for when x 332 m x le w

Answer 1

X	Y	XY	X²	Y²
0.332	2.8	0.9296	0.110224	7.84
0.276	7.7	2.1252	0.076176	59.29
0.34	4	1.36	0.1156	16
0.248	8.6	2.1328	0.061504	73.96
0.367	3.1	1.1377	0.134689	9.61
0.269	11.1	2.9859	0.072361	123.21

Ʃx =	1.832
Ʃy =	37.3
Ʃxy =	10.6712
Ʃx² =	0.570554
Ʃy² =	289.91
Sample size, n =	6
x̅ = Ʃx/n = 1.832/6 =	0.305333333
y̅ = Ʃy/n = 37.3/6 =	6.216666667
SSxx = Ʃx² - (Ʃx)²/n = 0.57055 - (1.832)²/6 =	0.011183333
SSyy = Ʃy² - (Ʃy)²/n = 289.91 - (37.3)²/6 =	58.02833333
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 10.6712 - (1.832)(37.3)/6 =	-0.71773333

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = -0.71773/√(0.01118*58.02833) = -0.8910

b)

Null and alternative hypothesis:

Ho: ρ = 0 ; Ha: ρ ≠ 0

Test statistic :  

t = r*√(n-2)/√(1-r²) = -0.891 *√(6 - 2)/√(1 - -0.891²) = -3.92

df = n-2 = 4

Critical value, t_c = T.INV.2T(0.1, 4) = 2.13

Conclusion:

Reject the null hypothesis. There is sufficient evidence.

d)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 58.02833 - (-0.71773)²/0.01118 = 11.96504232

Standard error, se = √(SSE/(n-2)) = √(11.96504/(6-2)) = 1.7295

Slope, b = SSxy/SSxx = -0.71773/0.01118 = -64.17884

y-intercept, a = y̅ -b* x̅ = 6.21667 - (-64.17884)*0.30533 = 25.812605

Regression equation :

ŷ = 25.8126 + (-64.1788) x

d)

Predicted value of y at x = 0.332

ŷ = 25.8126 + (-64.1788) * 0.332 = 4.5052

e)

Critical value, t_c = T.INV.2T(0.1, 4) = 2.1318

90% Confidence interval :

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 4.5052 - 2.1318*1.7295*√((1/6) + ((0.332 - 0.3053)²/(0.0112))) = 2.74

Upper limit = ŷ + tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 4.5052 + 2.1318*1.7295*√((1/6) + ((0.332 - 0.3053)²/(0.0112))) = 6.27

f)

Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

Test statistic:

t = b/(se/√SSxx) = -3.92

df = n-2 = 4

Critical value, t_c = T.INV.2T(0.1, 4) = 2.13

Conclusion:

Reject the null hypothesis. There is sufficient evidence.

g)

90% Confidence interval for slope:

Lower limit = β₁ - tc*se/√SSxx = -64.1788 - 2.1318*1.7295/√0.0112 = -99.04

Upper limit = β₁ + tc*se/√SSxx = -64.1788 + 2.1318*1.7295/√0.0112 = -29.31