In: Statistics and Probability
Let x be a random variable that represents the batting average of a professional baseball player. Let y be a random variable that represents the percentage of strikeouts of a professional baseball player. A random sample of n = 6 professional baseball players gave the following information.
x | 0.346 | 0.282 | 0.340 | 0.248 | 0.367 | 0.269 |
y | 3.3 | 7.4 | 4.0 | 8.6 | 3.1 | 11.1 |
(a) Verify that Σx = 1.852, Σy = 37.5, Σx2 = 0.583394, Σy2 = 288.43, Σxy = 10.845, and r ≈ -0.916.
Σx | |
Σy | |
Σx2 | |
Σy2 | |
Σxy | |
r |
(b) Use a 5% level of significance to test the claim that
ρ ≠ 0. (Use 2 decimal places.)
t | |
critical t ± |
Conclusion
Reject the null hypothesis, there is sufficient evidence that ρ differs from 0.
Reject the null hypothesis, there is insufficient evidence that ρ differs from 0.
Fail to reject the null hypothesis, there is insufficient evidence that ρ differs from 0.
Fail to reject the null hypothesis, there is sufficient evidence that ρ differs from 0.
(c) Verify that Se ≈ 1.4728, a ≈
25.438, and b ≈ -62.163.
Se | |
a | |
b |
(d) Find the predicted percentage of strikeouts for a
player with an x = 0.332 batting average. (Use 2 decimal
places.)
%
(e) Find a 95% confidence interval for y when x =
0.332. (Use 2 decimal places.)
lower limit | % |
upper limit | % |
(f) Use a 5% level of significance to test the claim that
β ≠ 0. (Use 2 decimal places.)
t | |
critical t ± |
Conclusion
Reject the null hypothesis, there is sufficient evidence that β differs from 0.
Reject the null hypothesis, there is insufficient evidence that β differs from 0.
Fail to reject the null hypothesis, there is insufficient evidence that β differs from 0.
Fail to reject the null hypothesis, there is sufficient evidence that β differs from 0.
(g) Find a 95% confidence interval for β and interpret its
meaning. (Use 2 decimal places.)
lower limit | |
upper limit |
Interpretation
For every unit increase in batting average, the percentage strikeouts increases by an amount that falls outside the confidence interval.
For every unit increase in batting average, the percentage strikeouts increases by an amount that falls within the confidence interval.
For every unit increase in batting average, the percentage strikeouts decreases by an amount that falls outside the confidence interval.
For every unit increase in batting average, the percentage strikeouts decreases by an amount that falls within the confidence interval.
a)
ΣX = | 1.852 |
ΣY= | 37.500 |
ΣX2 = | 0.583 |
ΣY2 = | 288.430 |
ΣXY = | 10.845 |
r = | -0.916 |
b)
test statistic t = | r*(√(n-2)/(1-r2))= | -4.57 | ||
t crit = | -/+ | 2.78 |
Reject the null hypothesis, there is sufficient evidence that ρ differs from 0.
c)
Se =√(SSE/(n-2))= | 1.4728 | |||
a= | 25.438 | |||
b= | -62.163 |
d)
predicted value = | 4.80 |
e)
std error of confidence interval = | s*√(1+1/n+(x0-x̅)2/Sxx)= | 1.6221 | |||||
for 95 % confidence and 4degree of freedom critical t= | 2.776 | ||||||
lower limit = | 0.30 | ||||||
uppr limit = | 9.30 |
f)
test statistic t = | r*(√(n-2)/(1-r2))= | -4.57 | ||
t crit = | -/+ | 2.78 |
Reject the null hypothesis, there is sufficient evidence that
β differs from 0.
g)
std error of slope sb1 = | s/√SSx= | 13.5909 | ||||
for 95 % confidence and -2degree of freedom critical t= | 2.7760 | |||||
95% confidence interval =b1 -/+ t*standard error= | (-99.891,-24.435) | |||||
lower limit = | -99.89 | |||||
uppr limit = | -24.44 |
For every unit increase in batting average, the percentage strikeouts increases by an amount that falls within the confidence interval.