STAT 511 - 001
Homework 12 Solution
Total 50
8.1
a.
Yes. It is an assertion about the value of a parameter.
b.
No. The sample median X is not a parameter.
c.
No. The sample standard deviation s is not a parameter.
d.
Yes. The assertion is that the standard deviation of population #2 exceeds that of
population #1
e.
No.
f.
Yes. H is an assertion about the value of a parameter.
a.
These hypotheses comply with our rules.
b.
Ha cannot include equality (i.e.
~
X and Y are statistics rather than parameters, so cannot appear in a hypothesis.
8.2
(1)
  20 ), so these hypotheses are not in compliance.
(1)
c.
d.
H0 should contain the equality claim, whereas Ha does here, so these are not legitimate.
(1)
The asserted value of 1
conditions are not met.
  2 in H0 should also appear in Ha. It does not here, so our
(1)
e.
Each S2 is a statistic, so does not belong in a hypothesis.
f.
We are not allowing both H0 and Ha to be equality claims (though this is allowed in more
comprehensive treatments of hypothesis testing).
(1)
g.
These hypotheses comply with our rules.
(1)
h.
These hypotheses comply with our rules.
(1)
(1)
8.7
A type I error here involves saying that the plant is not in compliance when in fact it is. A
type II error occurs when we conclude that the plant is in compliance when in fact it isn’t.
Reasonable people may disagree as to which of the two errors is more serious. If in your
judgement it is the type II error, then the reformulation H o :   150 v. H a :   150
makes the type I error more serious.
1
STAT 511 - 001
8.18
a.
72.3  75
 1.5 so 72.3 is 1.5 SD’s (of x ) below 75.
1 .8
b.
H0 is rejected if
c.
  area under standard normal curve below –2.88   2.88  .0020
(2)
d.
75  70 

  2.88 
   .1  .4602 so  70  .5398
9/5 

(2)
e.
 92.88  2.33 
n
 87.95 , so use n = 88
 75  70 
(2)
f.
Zero. By definition, a type I error can only occur when H 0 is true, but μ = 76 means that
H0 is actually false.
(2)
a.
Reject H0 if either
z  2.33 ; since z  1.5
is not
(1)
 2.33 , don’t reject H0.
(1)
2
8.19
z
z  2.58
or
z  2.58 ;

n
 0.3 , so
94.32  95
 2.27 . Since –2.27 is not < -2.58, don’t reject H0.
0.3


1 
1 

    2.58 
  5.91  .75  .2266
0.3 
0.3 

b.
 94   2.58 
c.
1.202.58  1.28 
n
  21.46 , so use n = 22.
95  94

a.
For n = 8, n – 1 = 7, and t .05, 7  1.895 , so H0 is rejected at level .05 if
2
8.29
Since
s
n

1.25
8
 .442 , t 
t  1.895 .
3.72  3.50
 .498 ; this does not exceed 1.895, so
.442
H0 is not rejected.
b.
d
o  
3.50  4.00

 .40 , and df = 7, so from table A.17,  .40  .73

1.25
2
STAT 511 - 001
8.30
n = 115, x  11.3 , s  6.43
1
Parameter of Interest:   true average dietary intake of zinc among males aged 65
– 74 years.
2
Null Hypothesis: H0:   15
3
Alternative Hypothesis: Ha:
4
z
5
6
x  o
s/ n

  15
x  15
(1)
(2)
s/ n
Rejection Region: No value of  was given, so select a reasonable level of
significance, such as = .05. z  z or z  1.645
z
(1)
11.3   o
6.43 / 115
(1)
 6.17
(2)
7
–6.17 < -1.645, so reject H0. The data does support the claim that average daily
intake of zinc for males aged 65 - 74 years falls below the recommended daily
allowance of 15 mg/day.
8.25(c)
 .32.58  2.33  
n
  216 .97 , so use n = 217.
 .1


2
8.37
1
2
3
p = true proportion of all donors with type A blood
H0: p = .40
Ha: p  .40
4
z
5
Reject H0 if z  2.58 or z  -2.58
6
z
7
pˆ  p o
p o 1  p o  / n
82 / 150  .40
.40.60 / 150


pˆ  .40
.40.60  / n
.147
 3.667
.04
Reject H0. The data does suggest that the percentage of all donors with type A blood
differs from 40%. (at the .01 significance level). Since the z critical value for a
significance level of .05 is less than that of .01, the conclusion would not change.
3
(2)
STAT 511 - 001
8.38
a.
Let X = the number of couples who lean more to the right when they kiss. If n = 124 and
p = 2/3, then E[X] = 124(2/3) = 82.667. The researchers observed x = 80, for a difference
of 2.667. The probability in question is
P(|X – 82.667| ≥ 2.667)
= P(X ≤ 80 or X ≥ 85.33)
= P(X ≤ 80) + [1 – P(X ≤ 85)]
= Bin(80;124,2/3) + [1 – Bin(85;124,2/3)]
= 0.634. (Using a large-sample z-based calculation gives a probability of 0.610.)
(3)
b.
We wish to test H0: p = 2/3 v. Ha: p ≠ 2/3.
(2)
80
 .645 , so our test statistic is
From the data, pˆ 
124
.645  .667
z
= –0.51.
(2)
.667 (.333 ) / 124
We would fail to reject H0 even at the α = .10 level, since |z| = 0.51 < 1.645. There is no
statistically significant evidence to suggest the p = 2/3 figure is implausible for rightleaning kissing behavior.
(2)
a.
p-value = .084 > .05 = , so don’t reject H0.
(1)
b.
p-value = .003 > .001 = , so don’t reject H0.
(1)
c.
.498 >> .05, so H0 can’t be rejected at level .05
(1)
d.
084 < .10, so reject H0 at level .10
(1)
e.
.039 is not < .01, so don’t reject H0.
(1)
f.
p-value = .218 > .10, so H0 cannot be rejected.
(1)
8.46
8.52
By guessing alone, the taster has a 1/3 chance of selecting the “different” wine.
Hence, we wish to test
H0: p= 1/3 v. Ha: p > 1/3.
346
With pˆ 
= .4047, our test statistic is
855
.4047  .3333
z
= 4.43
.3333 (.6667 ) / 855
and the corresponding P-value is
P(Z ≥ 4.43) ≈ 0.
4
(2)
(3)
(1)
STAT 511 - 001
Hence, we strongly reject the null hypothesis at any reasonable significance level and
conclude that the population of wine tasters have the ability to distinguish the “different” wine
out of three more than 1/3 of the time.
(2)
8.53
Here we might be concerned with departures above as well as below the specified weight of
s
5.0, so the relevant hypotheses are H0: μ = 5.0 v. Ha: μ ≠ 5.0. Since
 .035 ,
n
.13
z
 3.71 . Because 3.71 is “off” the z-table, P-value < 2(.0002) = .0004, so H0
.035
should be rejected.
5