Let x be a random variable that represents the percentage of successful free thr
ID: 3268682 • Letter: L
Question
Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of n = 6 professional basketball players gave the following information.
x 61 66 75 86 73 73 y 45 40 48 51 44 51 7. 10/22 points| Previous Answers BBUnderStat11 9.3.007 My Notes Ask Your Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of n = 6 professional basketball players gave the following information. 61 66 45 40 48 75 86 73 73 51 51 (a) Verify that Ex = 434, y = 279, x2 = 31756, = 13067, Exy = 20306, and r 0.678. Ex 434 Le2 31756 y213067 Exy 20306 r 0.678 (b) Use a 5% level of significance to test the claim that > 0, (Use 2 decimal places.) critical t Conclusion O Reject the null hypothesis, there is sufficient evidence that > 0 Reject the null hypothesis, there is insufficient evidence that > 0 O Fail to reject the null hypothesis, there is insufficient evidence that > 0 Fail to reject the null hypothesis, there is sufficient evidence that > 0 (c) Verify that Se3.5530, a21.615, b0.3440, and x72.333 Se 3.5530 a 21.615 b 0.3440 x 72.333Explanation / Answer
x
y
61
45
66
40
75
48
86
51
73
44
73
51
Question b)
t = r * sqrt ((n-2)/(1-r^2))
= 0.678 * sqrt ((6-2)/(1-0.678^2))
t = 1.84
If we use data analysis in excel then we get it as 1.85
From t-table we get one-tailed critical t at 5% level of significance for 4 degrees of freedom as 2.13
Answer:
t = 1.84
critical t = 2.13
Conclusion:
O Fail to reject the null hypothesis, there is insufficient evidence that r > 0.
Question d)
By using data analysis in excel we get the following output which gives us the line of regression.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.6782
R Square
0.4599
Adjusted R Square
0.3249
Standard Error
3.5530
Observations
6
ANOVA
df
SS
MS
F
Significance F
Regression
1
43.00458716
43.004587
3.407
0.138678664
Residual
4
50.49541284
12.623853
Total
5
93.5
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
21.6147
13.56065346
1.5939261
0.1861766
-16.03573102
59.26508882
X Variable 1
0.3440
0.186398901
1.85
0.1386787
-0.173489618
0.861563012
y^ = 21.6147 + 0.3340x
= 21.6147 + (0.3340*0.65)
= 43.32
If we use excel we get this value as 43.98
Answer: 43.32
Question e)
We use minitab here,
Predicted Values for New Observations
New
Obs Fit SE Fit 99% CI 99% PI
1 43.98 1.99 (34.80, 53.15) (25.22, 62.73)
Values of Predictors for New Observations
New
Obs x
1 65.0
Answer:
lower limit 25.2%
upper limit 62.7%
Question f)
This is same as part b
Answer:
t = 1.84
critical t = 2.13
If we use data analysis in excel then we get it as 1.85
Conclusion
Fail to reject the null hypothesis, there is insufficient evidence that b > 0.
x
y
61
45
66
40
75
48
86
51
73
44
73
51
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