Statistics 2080B: Practice Problems II

- 1.
Ten observations on , and

*Y*were stored in columns , and*C*_{4}. The MINITAB regression command**``regress c4 on 3 in C1 C2 C3"**was used to fit the modelA partial MINITAB regression output is included below.

*s*=9.074*R*^{2}=0.646 - a)
(The model is )

At least one of the above

*b*_{i}is not equal to zero.The model is

*F*=3.65.F has 3,6 df

*p*-valueReject

*H*_{0}at since the*p*-value is less than .The model is

- b)
(The model is )

At least one of the above

*b*_{i}is not equal to zero.The model is

F has 2,6 df

*p*-value =*P*(*F*_{2,6}> 1.07)*p*> 0.25Do not reject

*H*_{0}at since the*p*-value is greater than . - c)

- 2.
- a)
15
- b)
- c)
A male with no access to cable
.
A female with access to cable .

-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 dfFor CI,

*t*= 3.106.The confidence interval for is

(-7.848,0.774)

- e)
*l*'(*X*'*X*)^{-1}*l*= 0.252*t*has 11 dfFor PI,

*t*=2.201The 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*)^{-1}*l*= 0.7252*t*has 11 dfFor PI,

*t*=1.796The 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)

STATISTICS 2080 HOMEPAGE |