QUIZ 5 *** ANSWERS ***
#1 (5 points, 1 point each)
A
The phrase "...THE ABSENCE OF EVIDENCE IS NOT EVIDENCE OF ABSENCE..." refers to what
type of error ... Type I or Type II? Explain your answer.
TYPE II ERROR
YOU CAN NEVER PROVE A NULL HYPOTHESIS
-ORFAILURE TO REJECT THE NULL HYPOTHESIS CAN BE THE RESULT OF A TYPE II ERROR
(SHOULD SEE SOMETHING ABOUT NOT REJECTING THE NULL HYPOTHESIS)
B
Two researchers conduct a study. Both decide on the following prior to conducting the study:
alpha = .05; beta = .20. Researcher A conducts a study with 100 subjects. Researcher B
conducts a study with 1,000 subjects. Both studies result in a statistically significant finding.
Does either researcher have a greater probability of the significant finding being a result of
making a Type I error? Explain your answer.
BOTH HAVE THE SAME PROBABILITY OF A TYPE I ERROR SINCE BOTH STUDIES USE
ALPHA=0.05 (BOTH HAVE A 5% CHANCE OF MAKING A TYPE I ERROR)
Circle the correct answer ...
C
If you change the significance level of a hypothesis test from .05 to .01, the chance of a Type I
error is ...
higher
D
remains the same
If you change the significance level of a hypothesis test from .05 to .01, the chance of a Type II
error is ...
higher
E
lower
lower
remains the same
If you increase the sample size, the chance of a Type II error is ...
higher
lower
remains the same
#2 (6 points, 1 point each)
Ophthalmologists studying the treatment of using an infrared laser procedure in ten patients with vision
loss caused by dry age-related macular degeneration (AMD) found the following data on visual acuity
(VA) before and after the procedure ...
PERSON
TIME
1
2
3
4
5
6
7
8
9
10
LINES READ BEFORE LASER TREATMENT
3
4
2
4
6
3
5
5
3
3
LINES READ AFTER LASER TREATMENT
4
4
5
5
5
3
7
6
3
5
Use a α =.05 significance level to test the claim that there is a difference between the number of VA lines
that can be read by individuals before and after the procedure.
A
What are the null and alternative hypotheses?
B
What distribution would you use for hypothesis testing?
C
What is the critical value?
D
What is the value of the test statistic?
E
What is the p-value (your answer can be the values it is between rather than an exact value)?
extra credit 2 points ... explain what the p-value means
F
A
What is your conclusion?
H :µ = 0
0
d
H :µ ≠ 0
1
d
B
t-DISTRIBUTION
C
FROM t-TABLE WITH 9 DEGREES OF FREEDOM, 2-TAIL TEST,
ALPHA=0.05 (0.025 IN EACH TAIL)
2.262
D
t=
X
− 0.9
− 0.9
=
=
= −2.38
.
11972
s
0.3786
10
N
d
d
E
IN A t-TABLE, 2.38 IS BETWEEN 2.262 (0.025) AND 2.821 (.010)
TWO TAIL TEST, P-VALUE BETWEEN 0.05 AND 0.02 (EXACT VALUE IS 0.0414)
EXTRA CREDIT: IF THE NULL HYPOTHESIS IS TRUE, THERE IS A 4% CHANCE OF OBTAINING THE
RESULT OF A DIFFERENCE OF -0.9 BETWEEN THE BEFORE AND AFTER LINES READ
F
THERE IS A DIFFERENCE IN THE NUMBER OF LINES READ BEFORE AND AFTER THE LASER
TREATMENT
#3 (11 points)
Assume that you have identified a group of 100 men with high cholesterol levels. You have data on
other risk factors among these men such as age, smoking, weight, family history, etc. Based on other
research you have done, the combination of these other risk factors data predicts that 10% will have a
heart attack within 5 years. These men are followed for 5 years and 13% have heart attacks.
A (6 points)
Does this finding indicate anything about cholesterol and its effect on future heart
attacks? (HINT: 6 points ... hmmm ... look at A through F in question #2)
B (3 points)
Suppose you had 1000 men instead of 100 and also found that 13% have heart attacks.
Does this finding indicate anything about cholesterol and its effect on future heart
attacks? (HINT: 3 points ... hmmm ... look at D, E, F in question #2)
C (2 points)
Is there any difference between the conclusion you make with 100 versus 1000 men? If
so, why?
A1
H : p = 010
.
H : p ≠ 010
.
0
1
A2
NPQ = 100(0.1)(0.9)=9, NPQ>5, CAN USE NORMAL DISTRIBUTION (Z)
A3
FROM Z-TABLE, 2-TAIL TEST, ALPHA=0.05 (0.025 IN EACH TAIL)
1.96
A4
Z=
013
. − 010
.
0.03
p$ − p
=
=
=1
(01
. )(0.9) 0.03
pq
N
100
0
0
0
A5
FROM NORMAL TABLE, WITH Z=1, P=0.1587
TWO-TAIL TEST, EXACT P-VALUE=0.3174
A6
THERE IS NO EVIDENCE TO REJECT THE NULL HYPOTHESIS
B4
Z=
013
. − 010
.
0.03
p$ − p
.
=
=
= 316
(01
. )(0.9) 0.0095
pq
N
1000
0
0
0
B5
FROM NORMAL TABLE, WITH Z=3.16, P=0.0008
TWO-TAIL TEST, EXACT P-VALUE=0.0016
B6
TEST STATISTIC GREATER THAN THE CRITICAL VALUE, REJECT THE NULL HYPOTHESIS
THERE IS A DIFFERENCE IN HEART ATTACK FREQUENCY BETWEEN THIS GROUP AND THE
EXPECTED VALUE
C
THERE IS A DIFFERENT CONCLUSION. WITH A LARGER N, THERE IS MORE POWER TO DETECT
A DIFFERENCE (LESS OF A CHANCE OF MAKING A TYPE II ERROR)
#4 (6 points)
A study was conducted to compare indoor air quality in areas where smoking was permitted verus in
areas where smoking was not permitted. Measurements of carbon monoxide (CO) in parts per million
were made in 40 areas where smoking was permitted and in 40 areas where smoking was not permitted.
Results were as follows ...
AREA
N
MEAN CO
STANDARD DEVIATION CO
PERMITTED
40
11.6
7.3
NOT PERMITTED
40
8.5
3.5
Does CO concentration differ in the two areas? (HINT: 6 points ... hmmm ... look at A through F in
question #2)
A
H :µ = µ
0
1
2
H :µ ≠ µ
2
1
1
B
t-DISTRIBUTION
C
FROM t-TABLE WITH 40 DEGREES OF FREEDOM, 2-TAIL TEST,
ALPHA=0.05 (0.025 IN EACH TAIL)
2.021
*** REMEMBER, IN CLASS AND IN NOTES, I SAID TO USE SMALLER OF N1 AND N2 FOR DF ***
D
t=
116
31
X −X
. − 8.5
.
=
=
= 2.42
7.3 35
s s
.
128
.
+
+
40 40
n n
1
2
2
2
1
2
1
2
2
2
E
IN A t-TABLE, 2.42 IS BETWEEN 2.021 (0.025) AND 2.423 (.010)
TWO TAIL TEST, P-VALUE BETWEEN 0.05 AND 0.02
F
TEST STATISTIC GREATER THAN THE CRITICAL VALUE, REJECT THE NULL HYPOTHESIS
THERE IS A DIFFERENCE BETWEEN AREAS