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	67	65	75	86	73	73
y	42	40	48	51	44	51

(c) Verify that S_e ≈ 3.0468, a ≈ 8.188, b ≈ 0.5168, and x ≈ 73.167.

Se

=

(e) Find a 90% confidence interval for y when x = 83. (Round your answers to one decimal place.)

lower limit	%
upper limit	%

(f) Use a 5% level of significance to test the claim that β > 0. (Round your answers to two decimal places.)

t
critical t

Answer 1

Part c

Required regression model is given as below:

Simple Linear Regression Analysis
Regression Statistics
Multiple R	0.8139
R Square	0.6624
Adjusted R Square	0.5781
Standard Error	3.0468
Observations	6
ANOVA
	df	SS	MS	F	Significance F
Regression	1	72.8687	72.8687	7.8498	0.0487
Residual	4	37.1313	9.2828
Total	5	110.0000
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	8.188	13.5532	0.6041	0.5784	-29.4422	45.8172
x	0.5168	0.1845	2.8018	0.0487	0.0047	1.0289

From above regression model and descriptive statistics, we have

S_e ≈ 3.0468, a ≈ 8.188, b ≈ 0.5168, and x ≈ 73.167

Part e

Formula for confidence interval for mean response regression is given as below:

Y­_h -/+ t_{α/2, n– 2} *SE*sqrt[((1/n) + ((x – xbar)^2)/∑(x_i – xbar)^2))]

Where,

Y­_h = Predicted value of Y

df = n - 2

t_{α/2, n– 2} = Critical value

SE = standard error of estimate

n = sample size

Xbar = sample mean of x

We are given X = 83

Predicted value of Y = 8.188 + 0.5168*83

Predicted value of Y = 51.08186

Sample size = n = 6

df = 6 – 2 = 4

SE = 3.046774

Xbar = 73.16667

t_{α/2, n– 2} = 2.131847

(by using t-table)

∑(x_i – xbar)^2 = 272.8333

Confidence interval = Y­_h -/+ t_{α/2, n– 2} *SE*sqrt[((1/n) + ((x – xbar)^2)/∑(x_i – xbar)^2))]

Confidence interval = 51.08186 ± 2.131847*3.046774*sqrt[((1/6) + ((83 – 73.167)^2)/ 272.8333))]

Confidence interval = 51.08186 ± 4.6886

Lower limit = 51.08186 - 4.6886 = 46.3932

Upper limit = 51.08186 + 4.6886 = 55.77049

Lower limit = 46.4%

Upper limit = 55.8%

Part f

We are given α = 0.05

t = β/SE(β) = 0.5168/0.1845 = 2.8018

t = 2.80

Critical t value = 2.57