 # qnt/275 week 4 practice set, statistics for decision making Anonymous
timer Asked: Aug 6th, 2017
account_balance_wallet \$10

### Question Description

qnt/275 statistics for decision making, university of phoenix

jesusale932
School: Carnegie Mellon University   Here is my answer :)

Practice Set 4
QNT/275 Version 6

University of Phoenix Material
Practice Set 4
Practice Set 4
1. Find z for each of the following confidence levels. Round to two decimal places.
A. 90%
NormInv(

1−0.90

) = ±1.645

2

B. 95%
NormInv(

1−0.95

) = ±1.960

2

C. 96%
NormInv(

1−0.96

) = ±2.054

2

D. 97%
NormInv(

1−0.97

) = ±2.170

2

E. 98%
NormInv(

1−0.98

) = ±2.326

2

F. 99%
NormInv(

1−0.99
2

) = ±2.576

2. For a data set obtained from a random sample, n = 81 and x = 48.25. It is known
that σ = 4.8.
A. What is the point estimate of μ? Round to two decimal places
The point estimate μ is equal to 48.25
B. Make a 95% confidence interval for μ. What is the lower limit? Round to two decimal
places.

1

Practice Set 4
QNT/275 Version 6

x̅ − Zα/2

σ
√n

48.25 − 1.96 ∗

4.8
√81

48.25 − 1.05
𝐋𝐨𝐰𝐞𝐫 𝐥𝐢𝐦𝐢𝐭 = 𝟒𝟕. 𝟐𝟎
C. Make a 95% confidence interval for μ. What is the upper limit? Round to two decimal
places.
x̅ + Zα/2

σ
√n

48.25 + 1.96 ∗

4.8
√81

48.25 + 1.05
𝐔𝐩𝐩𝐞𝐫 𝐥𝐢𝐦𝐢𝐭 = 𝟒𝟗. 𝟑𝟎
D. What is the margin of error of estimate for part b? Round to two decimal places.

E = Zα/2

σ

√n
4.8
E = 1.96 ∗
√81
𝐄 = 𝟏. 𝟎𝟓
3. Determine the sample size (nfor the estimate of μ for the following.
A. E = 2.3, σ = 15.40, confidence level = 99%. Round to the nearest whole number.
α = 1 – 0.99 = 0.01
σ
n = (zα⁄2 ( ))2
E
15.40 2
n = (z0.01⁄2 (
))
2.3
15.40 2
n = (2.576 (
))
2.3
𝐧 = 𝟐𝟗𝟕

B. E = 4.1, σ = 23.45, confidence level = 95%. Round to the nearest whole number.

2

Practice Set 4
QNT/275 Version 6

α = 1 – 0.95 = 0.05
σ
n = (zα⁄2 ( ))2
E
15.40 2
n = (z0.05⁄2 (
))
2.3
23.45 2
n = (1.960 (
))
4.1
𝐧 = 𝟏𝟐𝟔
C. E = 25.9, σ = 122.25, confidence level = 90%. Round to the nearest whole number.
α = 1 – 0.90 = 0.10
σ
n = (zα⁄2 ( ))2
E
15.40 2
n = (z0.10⁄2 (
))
2.3
122.25 2
n = (1.645 (
))
25.9
𝐧 = 𝟔𝟎
4. True or False.
a. The null hypothesis is a claim about a population parameter that is assumed to be false
until it is declared false.
A. True
B. False
b. An alternative hypothesis is a claim about a population parameter that will be true if the
null hypothesis is false.
A. True
B. False
c. The critical point(s) divide(s) is some of the area under a distribution curve into rejection
and nonrejection regions.
A. True
B. False
d. The significance level, denoted by α, is the probability of making a Type II error, that is,
the probability of rejecting the null hypothesis when it is actually true.
A. True
B. False
e. The nonrejection region is the area to the right or left of the critical point where the null
hypothesis is not rejected.
A. True

3

Practice Set 4
QNT/275 Version 6

B. False
5. Fill in the blank. The level of significance in a test of hypothesis is the probability of making a
_Type I error_. It is the area under the probability distribution curve where we reject H0.
A. Type I error
B. Type II error
C. Type III error
6. Consider H0: μ = 45 versus H1: μ < 45. A random sample of 25 observations produced a sample
mean of 41.8. Using α = .025 and the population is known to be normally distributed with σ = 6.
A. What is the value of z? Round to two decimal places.
x−μ
σ
( )
√n
41.8 − 45
Z=
6
(
)
√25
𝐙 = −𝟐. 𝟔𝟕
Z=

B. Would you reject the null hypothesis?

For α = .025 the critical value is -1.96. This value is greater than -2.67, then reject the
null hypothesis.
1. Reject Ho
2. Do not reject Ho
7. The following information is obtained from two independent samples selected from two
normally distributed populations.
n1 = 18

x1 = 7.82

σ1 = 2.35

n2 =15

x2 =5.99

σ2 =3.17

A. What is the point estimate of μ1 − μ2? Round to two decimal places.
μ1 − μ2 = 7.82 – 5.99 = 1.83
B. Construct a 99% confidence interval for μ1 − μ2. Find the margin of error for this
estimate.
Round to two decimal places.

Copyright © 2017 by University of Phoenix. All righ...

flag Report DMCA  Review Anonymous
Thank you! Reasonably priced given the quality not just of the tutors but the moderators too. They were helpful and accommodating given my needs. Brown University

1271 Tutors California Institute of Technology

2131 Tutors Carnegie Mellon University

982 Tutors Columbia University

1256 Tutors Dartmouth University

2113 Tutors Emory University

2279 Tutors Harvard University

599 Tutors Massachusetts Institute of Technology

2319 Tutors New York University

1645 Tutors Notre Dam University

1911 Tutors Oklahoma University

2122 Tutors Pennsylvania State University

932 Tutors Princeton University

1211 Tutors Stanford University

983 Tutors University of California

1282 Tutors Oxford University

123 Tutors Yale University

2325 Tutors