Statistics 2080B: Practice Problems II

1.

Ten observations on , and Y were stored in columns , and C4. The MINITAB regression command regress c4 on 3 in C1 C2 C3" was used to fit the model

A partial MINITAB regression output is included below.

s=9.074

R2=0.646

a)

(The model is )

At least one of the above bi is not equal to zero.

The model is

F=3.65.

F has 3,6 df

p-value

Reject H0 at since the p-value is less than .

The model is

b)

(The model is )

At least one of the above bi is not equal to zero.

The model is

F has 2,6 df

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

p > 0.25

Do not reject H0 at since the p-value is greater than .

c)

2.

a) 15

b)

-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)

t has 15 - 4 = 11 df

For CI, t = 3.106.

The confidence interval for is

(-7.848,0.774)

e)

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

t has 11 df

For PI, t=2.201

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

(7.71 hours per week,20.07 hours per week)

f)

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

t has 11 df

For PI, t=1.796

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

(9.90 hours per week,17.75 hours per week)

Jonathan Payne
1999-03-11