Statistics 2080B: Practice Problems II

1.

Ten observations on $X_{1}, \ X_{2}, \ X_{3}$, and Y were stored in columns $C_{1}, \ C_{2}, \ C_{3}$, and C4. The MINITAB regression command ``regress c4 on 3 in C1 C2 C3" was used to fit the model

$y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \varepsilon$

A partial MINITAB regression output is included below.

s=9.074

R2=0.646

$\begin{array}{lcccc} & & Analysis of Variance & & \\ Source & DF & SS & MS & ... ... & 3.65 \\ Error & 6 & 494.07 & 82.35 & \\ Total & & 1396.02 & & \end{array}$

$\begin{array}{lcc} Source & DF & SEQSS \\ C_{1} & 1 & 726.22 \\ C_{2} & 1 & 35.48 \\ C_{3} & 1 & 140.25 \end{array}$

a)

$H_{0}: \ \ \ \ \beta_{1} = \beta_{2} = \beta_{3} = 0$ (The model is $y = \beta_{0} + \varepsilon$)

$H_{A}: \ \ \ $ At least one of the above bi is not equal to zero.

The model is $y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \varepsilon$

F=3.65.

F has 3,6 df

p-value $= P(F_{3,6} > 3.65) \ \ \ \ 0.05 < p < 0.10$

Reject H0 at $\alpha = 0.10$ since the p-value is less than $\alpha$.

The model is $y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \varepsilon$

b)

$H_{0}: \ \ \ \ \beta_{2} = \beta_{3} = 0$ (The model is $y = \beta_{0} + \beta_{1}x_{1} + \varepsilon$)

$H_{A}: \ \ \ $ At least one of the above bi is not equal to zero.

The model is $y = \beta_{0} + \beta_{1}x_{1} + \beta_{2}x_{2} + \beta_{3}x_{3} + \varepsilon$

$F = \frac{\frac{SS_{drop}}{k - g}}{MSE_{full}} = 1.07$

F has 2,6 df

p-value = P(F2,6 > 1.07)

p > 0.25

Do not reject H0 at $\alpha = 0.10$ since the p-value is greater than $\alpha$.

c)

$y = \beta_{0} + \beta_{1}x_{1} + \varepsilon$


2.

a) 15

b) $\hat{y} = 8.951 + 0.247x_{1} - 3.537x_{2} + 3.468x_{3}$

c) A male with no access to cable $x_{2}=0, \ x_{3}=0$.

A female with access to cable $x_{2}=1, \ x_{3}=1$.

-3.537 + 3.468 = -0.069 hours

All else being equal, a male with no access to cable watches on average 0.069 hours more television per week than a female with access to cable.

d) $\hat{\beta}_{2} = -3.537$

$s_{\hat{\beta}_{2}} = s\sqrt{v_{22}} = 2.51\sqrt{0.3058} = 1.388$

t has 15 - 4 = 11 df

For $99\%$ CI, t = 3.106.

The $99\%$ confidence interval for $\beta_{2}$ is $-3.537 \pm 3.106(1.388)$

$-3.537 \pm 4.311$

(-7.848,0.774)

e)

$l=\left(\begin{array}{c} 1 \\ 20 \\ 0 \\ 0 \end{array}\right)$

l'(X'X)-1l = 0.252

t has 11 df

For $95\%$ PI, t=2.201

$\hat{y} = 8.951 + 0.247(20) - 3.537(0) + 3.468(0) = 13.891$

The $95\%$ prediction interval for the amount of television watched per week by a 20 year old male with no access to cable is

$\hat{y} \pm ts\sqrt{1 + l'(X'X)^{-1}l}$

$13.891 \pm 2.201(2.51)\sqrt{1 + 0.252}$

$13.891 \pm 6.182$

(7.71 hours per week,20.07 hours per week)

f)

$l=\left(\begin{array}{c} 1 \\ 20 \\ 1 \\ 1 \end{array}\right)$

l'(X'X)-1l = 0.7252

t has 11 df

For $90\%$ PI, t=1.796

$\hat{y} = 8.951 + 0.247(20) - 3.537(1) + 3.468(1) = 13.822$

The $90\%$ confidence interval for the amount of television watched per week by a 20 year old female with access to cable is

$\hat{y} \pm ts\sqrt{l'(X'X)^{-1}l}$

$13.822 \pm 1.796(2.51)\sqrt{0.7572}$

$13.822 \pm 3.923$

(9.90 hours per week,17.75 hours per week)


STATISTICS 2080 HOMEPAGE
Jonathan Payne
1999-03-11