Transcript Answer KEY - Day 2 of Prop Test Worksheet

Homework
Worksheet on one Sample
Proportion Hypothesis Tests
A drug company has developed a new vaccine for preventing the flu. The
company claims that fewer than 5% of adults who use its vaccine will get
the flu.
1. State the hypotheses.
2. Describe a Type I error and it’s consequence
3. Describe a Type II error and it’s consequence
4. Should I use a level of significance of 0.01, 0.05, or 0.1?
#1
In a poll, 432 of 1004 adults surveyed said that they
believe the Census Bureau when it says the
information you give is kept confidential. Is there
convincing evidence that fewer than half of U.S. adults
believe the information is kept confidential. Use a 0.01
significance level.
In a poll, 432 of 1004 adults surveyed said that they believe the Census Bureau when it says
the information you give is kept confidential. Is there convincing evidence that fewer than half
of U.S. adults believe the information is kept confidential. Use a 0.01 significance level.
p = Pop. Prop. of adults who believe the Census information is kept confidential.
Ho : p  0.50

H A : p  0.50
z
z
Assumptions:
1. SRS
2. Approx. Norma
4.44
np  1004(.5)  502  5

nq  1004(.5)  502  5
pp
pq
n
0.43  0.50
0.5(0.5)
1004
z  4.44
P  val  P ( p  0.43)  P ( z  4.44)
ncdf ( , 4.44)
3. 10(1004)=10040
(Independent
Pval  4.5 x10 6  0
Reject the Ho since P-value (0)<α (0.01).
There is sufficient evidence to support the claim that less than
half of Americans believe that the Census information is kept
confidential.
#2
APL patients were given an arsenic compound as part of their treatment. Of those receiving
the arsenic, 42% were in remission and showed no signs of leukemia in later exams. It is
know that 15% of APL patients go into remission after conventional treatment. Suppose the
study included 100 randomly selected patients. Is there sufficient evidence to conclude that
the proportion in remission is greater than 0.15? Use a 0.01 significance level
.
.
Appoximately 42% of the 100 randomly selected APL patient who received
arsenic were in remission. We know that 15% of APL patients go into remission
after conventional treatment. Is there sufficient evidence to conclude that the
proportion in remission is greater than 0.15? Use a 0.01.
p = Pop. Prop. of APL patients who go into remission.
Assumptions:
Ho : p  0.15

H A : p  0.15
1. SRS
2. Approx. Normal
np  100(.15)  15  5

nq  100(.85)  85  5
7.56
z
z
pp
pq
n
0.42  0.15
0.15(0.85)
100
z  7.56
P  val  P ( p  0.42)  P ( z  7.56)
3. 10(100) = 1000
ncdf (7.56,  )
Pval  2.03 x10 14  0
Independent
Reject the Ho since P-value (0)<α (0.01).
There is sufficient evidence to support the claim that the
proportion in remission is greater than 15%.
#3
The Associated Press reported that 715 of Americans, age
25 and older, are overweight, a substantial increase over
the 58% figure from 1983. Although this information came
from a Harris Poll rather than a census of the population,
let’s assume for the purp0ses of this exercise that the
nationwide population proportion is exactly 0.71. suppose
that an investigator wishes to know whether the proportion
of such individuals in her state who are overweight differs
from the national proportion. A random sample of size n =
600 results in 450 who are classified as overweight.
a. What can the investigator conclude? Answer this
question by carrying out a hypothesis test with a
1% significance level.
b. Describe Type I and Type II error
c. Which type of error might you be making?
The population proportion is exactly 0.71. An investigator want to check to see if
her state’s proportion is different. Out of 600, 450 are overweight. Use 1%
significance level.
p = Pop. Prop. of who are considered overweight.
Ho : p  0.71

H A : p  0.71
z
z
pp
pq
n
0.75  0.71
0.71(0.29)
600
z  2.16
Assumptions:
1. SRS
2.16
2. Approx. Normal
np  600(.71)  426  5

nq  600(.29)  174  5
P  val  P ( p  .75) * 2  P ( z  2.16) * 2
ncdf (2.16,  ) * 2
3. 10(600)=6000
Pval  0.030
Independent
Fail to Reject the Ho since P-value (0.03)>α (0.01).
There is insufficient evidence to support the claim that
the proportion of overweight people is different from
nationwide proportion of 71%
#4
Georgia’s HOPE scholarship program guarantees fully paid
tuition to Georgia public universities for Georgia high school
seniors who have a B averrage in academic requirements as
long as they maintain a B average in college. Of 137 randomly
selected students enrolling in the Ivan Allen College at the
Georgia Institute of Technology in 1996 who had a B average
going into colllege, 53.2% had a GPA below 3.0 at the end of
their first year. Do these data provide convincing evidence that
a majority of students at Ivan Allen College who enroll with a
HOPE scholarship lose their scholarship?
Of 137 randomly selected students at Ivan, 53.2% had a GPA below 3.0 at the end of
.
their first year. Do these data provide convincing evidence that a majority of students
at Ivan Allen College who enroll with a HOPE scholarship lose their scholarship?
p = Pop. Prop. of Ivan College students who lose their scholarship
Assumptions:
Ho : p  0.5

H A : p  0.5
1. SRS
2. Approx. Normal
0.75
z
z
pp
pq
n
0.532  0.5
0.5(0.5)
137
z  0.75
P  val  P ( p  0.532)  P ( z  0.75)
ncdf (0.75,  )
Pval  0.2266
np  137(.5)  68.5  5

nq  137(.5)  68.5  5
3. 10(137)=1370
Independent
Fail to Reject the Ho since P-value (0.2266)>α (0.05).
There is insufficient evidence to support the claim that
the proportion of Ivan College students who lose their
scholarship is more than 50%.
#5
Teenagers (age 15 to 20) make up 7% of the driving
population. A study of auto accidents conducted by the
Insurance Institute for Highway Safety found that 14%
of the accidents studied involved teenage drivers.
Suppose that this percentage was based on examining
records from 500 randomly selected accidents. Does
the study provide convincing evidence that the
proportion of accidents involving teenage drivers is
greater than 7%, the proportion of teenage drivers?
.
Sample of 500 had 14% of accidents had teenage drivers. Test at 5 % significance
against the population proportion of teenage drivers which is 7%.
p = Pop. Prop of auto accidents involving teenage drivers
Assumptions:
1. SRS
2. Approx. Norma
3. 10(500)=5000 Independent
Reject the Ho since P-value (0)< α (0.05).
There is sufficient evidence to support the claim that
the proportion of teenage drivers involved in accidents
is greater than the proportion of teenage drivers (7%).
#6
Students at the Akademia Podlaka conducted an experiment to
determine whether the Belgium-minted Euro coin was equally
likely to land heads up or tails up. Coins were spun on a smooth
surface, and in 250 spins, 140 landed with the heads side up.
Test this using a 1% significance level.
.
140 out of 250 land on heads up. Is it equally likely. Test using a 1% significance
level.
p = Pop. Prop of auto accidents involving teenage drivers
Assumptions:
1. SRS
2. Approx. Norma
3. 10(500)=5000 Independent
Fail to Reject the Ho since P-value (0.0574)> α (0.01).
There is insufficient evidence to support the claim that
the coin is not equally likely to land on heads as tails.