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STAT 420 Spring 2009
Homework
(10 points) (due Friday, January 30, by 3:00 p.m.)
From the textbook:
The salary of junior executives in a large retailing firm is normally distributed with
standard deviation σ = $1,500. If a random sample of 25 junior executives yields
an average salary of $16,400, what is the 95-percent confidence interval for μ, the
average salary of all junior executives.
σ = $1,500, n = 25, X = $16,400.
n
z σ
2
X± α
95% confidence, α = 0.05, z (^) 0.025 = 1.960.
16 , 400 ± 1. 960 ⋅^1 ,^500 16,400 ±±±± 588 ( 15,812 , 16,988 )
Redo Problem 3.5 without assuming that the population standard deviation is
known. You are given s = $1,575.
s = $1,575, n = 25, X = $16,400. ( )
n
t n 1 s
2
X ± α −
95% confidence, α = 0.05, t (^) 0.025 ( 24 ) = 2.064.
16 , 400 ± 2. 064 ⋅^1 ,^575 16,400 ±±±± 650.16 ( 15,749.84 , 17,050.16 )
3.18 (a), (b)
A cereal packaging machine is supposed to turn out boxes that contain 20 oz of cereal on the average. Past experience indicates that the standard deviation of the
content weights of the packages turned out by the machine is σ = 0.25 oz. Boxes
containing less than 19.8 oz of cereal, on the average, are considered unacceptable.
Thus the manufacturer wishes to test H 0 : μ = 20 and H a : μ = 19.8.
Hint: Treat this problem as a left-tailed test since 19.8 < 20. Also assume that the cereal weights are approximately normally distributed.
a) If samples of size n = 20 are drawn from the current production to test the
hypotheses, fund the power function of the test assuming a significance level
of α = 0.025.
Hint: Recall: Power = P ( Reject H 0 | H 0 is not true ) = P ( Reject H 0 | μ = 19.8 ).
Find the rejection region first and state it in terms of the sample mean X.
n = 20. α = 0.025.
Rejection Region:
Z =
n
X − μ 0 < – z
α.^ Z^ =
X − 20 < – 1.96.
X < 20 − 1. 96 ⋅^0.^25 = 19.89.
Power = P ( X < 19.89 | μ = 19.8 ) =
P Z^19.^8919.^8
= P ( Z < 1.61 ) = 0..
3.30 (a), (b)
Suppose an experiment yields m = 9, n = 12, X 1 = 68, X 2 = 60, s^21 = 25, and
s^22 = 30.
( Assume that the two populations are approximately normally distributed, and that
the overall variances are equal. )
a) Use the two-sample t statistic to test H 0 : μ 2 – μ 1 = 0 against H a : μ 2 – μ 1 ≠ 0
at a significance level of your choice.
pooled (^) + −
= −^ ⋅ + − ⋅
m n
s m s n^ s = (^ )^ (^ )
Test Statistic: t = (^ )^ (^ )
27.895^1
pooled
m n
s
x x = 3.435.
Rejection Region:
Reject H 0 if t < – t 0.025 ( m + n – 2 = 19 d.f. ) OR t > t 0.025 ( 19 d.f. )
t 0.025 ( 19 ) = 2.093. Reject H 0 at αααα = 0.05.
OR
3.435 > 2.861 (^) 2 – tailed
t >^ t 0.005 ( 19 ) p-value < 0.005^ ×^ 2 = 0.01.
Reject H 0 at αααα = 0.01.
b) Construct 90- and 95-percent confidence intervals for μ 2 – μ 1.
− ± ⋅ ⋅^ +
m n
x 2 x 1 t α 2 s pooled^211 m + n – 2 = 19 d.f.
− ± ⋅ ⋅^ +
60 68 1. 729 27.895^1
− ± ⋅ ⋅^ +
60 68 2. 093 27.895^1
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
2. For a group of 12 men (35-40 years old) who did not exercise regularly, the average
blood cholesterol was 270, and the sample standard deviation was 37. For a group of 7 men (also 35-40 years old) who did exercise regularly, the average blood cholesterol was 220, with the sample standard deviation of 31. We wish to test the claim that regular exercise lowers cholesterol by more than 20 points, on average. What is the
p- value for the test H 0 : μ NE – μ E ≤ 20 vs. H a : μ NE – μ E > 20? ( Assume that
the two populations are approximately normally distributed, and that the overall
variances are equal. )
pooled (^) + −
s = − ⋅ + − ⋅ = 1225
s pooled= 35
Test Statistic: T = (^ )^ (^ )
X Y
1 2 pooled
n n
s
n 1 + n 2 – 2 = 12 + 7 – 2 = 17 d.f.
0.025 < p-value ( right – tailed ) < 0.05 ( p-value ≈ 0.0446 )
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
From the textbook:
Prove the algebraic identity
∑ ∑ (^ )^ ∑ (^ )^ ∑ ∑(^ )
= = = = =
J j
ij j
J j
j
J j
ij
n i
Y Y Y Y
n i
Y Y n
1 1
2 1
2 1 1
2
∑ ∑ (^ )^ ∑ ∑(^ )
= = = =
J j
n
i
ij j j
J j
n
i
ij
j j Y Y Y Y Y Y 1 1
2 1 1
2
= = = = = =
J j
n
i
j
J j
n
i
ij j j
J j
n
i
ij j
j j j Y Y Y Y Y Y Y Y 1 1
2 1 1 1 1
= = = = =
J j
j j
J j
n
i
j ij j
J j
n
i
Yij Yj Y Y Y Y n Y Y
j j 1
2 1 1 1 1
Since ∑( )
=
n j
i
Yij Yj 1
= 0, j = 1, … , J , ∑ ( ) ∑( )
= =
J j
n
i
j ij j
j Y Y Y Y 1 1
2 = 0, and
∑ ∑ (^ )^ ∑ ∑(^ )^ ∑ (^ )
= = = = =
J j
j j
J j
n
i
ij j
J j
n
i
Yij Y Y Y n Y Y
j j 1
2 1 1
2 1 1
OR
1 1 1 1
2 1 1
2 2 1 1
Y Y^2 Y 2 Y Y Y Y 2 Y^ J Y JnY j
n i
ij
J j
n i
ij
J j
n i
ij ij
J j
n i
∑ ∑ ij −^ =∑ ∑^ − + =∑ ∑ − ∑ ∑ +
= = = = = = = =
1 1
(^22) 1 1
Y^2 2 Y JnY JnY^ J Y JnY j
n i
ij
J j
n i
∑ ∑ ij −^ + =∑ ∑ −
= = = =
1 1
2 1
2 2 1
nY Y^2 n Y 2 Y Y Y nY 2 Y^ J nY JnY j
j
J j
j
J j
j j
J j
∑ j −^ =∑ ^ − + =∑ − ∑ +
= = = =
1
2 2 1
nY^2 2 Y n JY JnY^ J nY JnY j
j
J j
∑ j −^ + =∑ −
= =
ANOVA table. (You should be able to calculate easily all the ANOVA table entries
given the following summary measures: J = 3, n = 6, Y A = 70, YB = 60, YC = 62,
2
s A = 78.8,
2
s B = 169.2,
2
s C = 140.0.)
N = n A + n B + n C = 6 + 6 + 6 = 18.
Y = 6 ⋅^70 +^6 ⋅^60 +^6 ⋅^62 = = 64.
SSB = n A ⋅( YA − Y ) 2 + nB ⋅( YB − Y ) 2 + nC ⋅( YC − Y ) 2
MSB =
SSB =
J −
SSW = ( nA − 1 ) ⋅ sA^2 +( nB − 1 )⋅ sB^2 +( nC − 1 ) ⋅ sC^2
MSW =
SSW = 1940
N − J
SSTot = SSB + SSW = 336 + 1940 = 2276.
F =
MSW
MSB ≈ ≈ 1.29897.
ANOVA table:
Source SS DF MS F
Between^336 2 168 1.
Within^1940 15 129.
Total^2276
F 0.05 ( 2, 15 ) = 3.68. Do NOT Reject H 0 : μ A = μ B = μ C at αααα = 0.05.
p-value = [ = FDIST ( 1.29897 , 2 , 15 ) ] = 0.3018.
