wK 6 (1)
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American Public University *
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Subject
Statistics
Date
Apr 26, 2024
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Uploaded by CountFireDinosaur40
Week 6 Knowledge Check Homework Practice Questions - Re…
Attempt 1 of 4
Written Apr 12, 2024 12:07 AM - Apr 12, 2024 2:47 AM
Attempt Score
15 / 20 - 75 %
Overall Grade (Highest Attempt)
15 / 20 - 75 %
Question 1
1 / 1 point
A manager wishes to see if the time (in minutes) it takes for their workers to complete a certain task is faster if
they are wearing earbuds. A random sample of 20 workers' times were collected before and after wearing
earbuds. Test the claim that the time to complete the task will be faster, i.e. meaning has production increased,
at a significance level of α = 0.01
For the context of this problem, μ
D
= μ
before
−μ
after
where the first data set represents before earbuds and the
second data set represents the after earbuds. Assume the population is normally distributed. The hypotheses are:
H
0
: μ
D
= 0
H
1
: μ
D
> 0
You obtain the following sample data:
Before
After
68
62.3
72.5
61.6
39.3
21.4
67.7
60.4
38.3
47.9
85.9
78.6
67.3
75.1
Before
After
59.8
48.3
72.1
65
79
83
61.7
56.8
57.9
44.7
56.8
50.6
71
63.4
80.6
68.9
59.8
33.9
73.1
79
49.9
38.4
59.2
55.4
64.8
55.6
Choose the correct decision, summary and state the p-value.
Hide ques±on 1 feedback
Reject H
0
, there is enough evidence to support the claim that the time to complete the task has
decreased when workers are allowed to wear earbuds at work and the p-value = .0024
Reject H
0
, there is enough evidence to support the claim that the time to complete the task has
decreased when workers are allowed to wear earbuds at work and the p-value = .0012
Do not reject H
0
, there is not enough evidence to support the claim that the time to complete the task
has decreased when workers are allowed to wear earbuds at work and the p-value = .0024.
Do not reject H
0
, there is enough evidence to support the claim that the time to complete the task has
decreased when workers are allowed to wear earbuds at work and the p-value = .0012
Copy and paste the data into Excel. Use the Data Analysis Toolpak in Excel.
Data - > Data Analysis -> scroll to where is says t:Test: Paired Two Samples for Means -
> OK
Question 2
1 / 1 point
A physician wants to see if there was a difference in the average smokers' daily cigarette consumption after
wearing a nicotine patch. The physician set up a study to track daily smoking consumption. In the study, the
patients were given a placebo patch that did not contain nicotine for 4 weeks, then a nicotine patch for the
following 4 weeks. Test to see if there was a difference in the average smoker's daily cigarette consumption
using α = 0.01. The hypotheses are:
H
0
: μ
D = 0
H
1
: μ
D ≠ 0
t-Test: Paired Two Sample for Means
Placebo
Nicotine
Mean
16.75
10.3125
Variance
64.46667 33.29583
Observations
16
16
Pearson Correlation
0.6105
Hypothesized Mean Difference
0
df
15
Variable 1 Range: is Before
Variable 2 Range: is After
The Hypothesized Mean Difference is 0 and make sure you click Labels in the first row
and click OK. You will get an output and this is the p-value you are looking for.
P(T<=t) one-tail 0.0012
t Stat
4.0119
P(T<=t) one-tail
0.0006
t Critical one-tail
2.6025
P(T<=t) two-tail
0.0011
t Critical two-tail
2.9467
What is the correct decision?
Hide ques±on 2 feedback
Question 3
1 / 1 point
A researcher took a random sample of 200 new mothers in the United States (Group 1) and found that 16% of
them experienced some form of postpartum depression. Another random sample of 200 new mothers in France
(Group 2), where mothers receive 16 weeks of paid maternity leave, found that 11% experienced some form of
postpartum depression. Can it be concluded that the percent of women in the United States experience more
postpartum depression than the percent of women in France? Use α=0.05
.
Select the correct alternative hypothesis and decision
.
Accept H
1
Accept H
0
Reject H
1
Reject H
0
Do not reject H
0
The p-value for a two tailed test is 0.0011. This is given to you in the output. No
calculations are needed. 0.0011 < .01, Reject Ho, this is significant.
H
1
: p
1
≠ p
2
; Do not reject the null hypothesis.
H
1
: p
1
≠ p
2
; Reject the null hypothesis.
Hide ques±on 3 feedback
Question 4
0 / 1 point
___
-0.291___
(-0.356, -.356)
___
-0.225___
(0.139, .139)
Hide ques±on 4 feedback
H
1
: p
1
> p
2
; Do not reject the null hypothesis.
H
1
: p
1
> p
2
; Reject the null hypothesis.
H
1
: p
1
< p
2
; Do not reject the null hypothesis.
H
1
: p
1
< p
2
; Reject the null hypothesis.
This is an upper tailed test because because of the keyword more.
z = z = 1.463171
Use NORM.S.DIST(1.463171,TRUE) to find the for the lower tailed test.
0.92829, then 1 - 0.92829, this is the p-value you want to use for an upper tailed test.
Two competing toothpaste brands both claim to produce the best toothpaste for whitening. A dentist randomly
samples 48 patients that use Brand A (Group 1) and finds 30 of them are satisfied with the whitening results of
the toothpaste. She then randomly samples 45 patients that use Brand B (Group 2) and finds 33 of them are
satisfied with the whitening results of the toothpaste. Construct a 99% confidence interval for the difference in
proportions and use it to decide if there is a significant difference in the satisfaction level of patients.
Enter the confidence interval - round to 3 decimal places.
< p
1
- p
2
< Z-Critical Value =NORM.S.INV(.995) = 2.575
Question 5
1 / 1 point
___
0.2958___
Hide ques±on 5 feedback
LL = (.625-.7333) - 2.575*
UL = (.625-.7333) + 2.575*
A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and
perform well on their test. During finals week, she randomly samples 45 students and asks them at the door
what they ate for breakfast. She categorizes 25 students into Group 1 as those who ate a healthy breakfast that
morning and 20 students into Group 2 as those who did not. After grading the final, she finds that 48% of the
students in Group 1 earned an 80% or higher on the test, and 40% of the students in Group 2 earned an 80% or
higher. Can it be concluded that eating a healthy breakfast improves test scores? Use a 0.05 level of
significance.
Hypotheses:
H
0
: p
1
= p
2
H
1
: p
1
> p
2
In this scenario, what is the p-value? (Round to 4 decimal places)
p-value =
z =
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