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
...
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, μD = μbefore−μafter 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:
H0: μD = 0
H1: μD > 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
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.
A.
Do not reject H0, 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
B.
Do not reject H0, 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.
C.
Reject H0, 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
D.
Reject H0, 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
Answer Key:D
Feedback:
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
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
Question 2 of 20
1.0/ 1.0 Points
A researcher is testing reaction times between the dominant and non-dominant hand. They
randomly start with each hand for 20 subjects and their reaction times in milliseconds are
recorded. Test to see if the reaction time is faster for the dominant hand using a 5% level of
significance. The hypotheses are:
H0 : μD = 0
H1 : μD > 0
t-Test: Paired Two Sample for Means
NonDominant Dominant
Mean 63.33 56.28
Variance
218.96431
58
128.75221
05
Observations 20 20
Pearson Correlation 0.9067
Hypothesized Mean
Difference 0
df 19
t Stat 4.7951
P(T<=t) one-tail 0.0001
t Critical one-tail 1.7291
P(T<=t) two-tail 0.0001
t Critical two-tail 2.0930
What is the correct decision?
A.
Do not reject H0
B.
Reject H1
C.
Accept H0
D.
Accept H1
E.
Reject H0
Answer Key:E
Feedback:The p-value for a one tailed test is 0.0001. This is given to you in the output. No
calculations are needed. 0.0001 < .05, Reject Ho, this is significant.
Part 2 of 7 - Week 6: Comparing Two Independent Population Proportions 2.5/ 3.0 Points
Question 3 of 20
1.0/ 1.0 Points
TDaP is a booster shot that prevents Diphtheria, Tetanus, and Pertussis in adults and adolescents.
It should be administered every 8 years for it to remain effective. A random sample of 500
people living in a town that experienced a pertussis outbreak this year were divided into two
groups. Group 1 was made up of 132 individuals who had not had the TDaP booster in the past 8
years, and Group 2 consisted of 368 individuals who had. In Group 1, 15 individuals caught
pertussis during the outbreak, and in Group 2, 11 individuals caught pertussis. Is there evidence
to suggest that the proportion of individuals who caught pertussis and were not up to date on
their booster shot is higher than those that were? Test at the 0.05 level of significance.
Enter the P-Value - round to 4 decimal places.
p-value = 0.0001
Answer Key:0.0001|.0001
Feedback:
This is an upper tailed test because because of the keyword higher.
z =
z = 3.717742
Use NORM.S.DIST(3.717742,TRUE) to find the for the lower tailed test.
0.999899, then 1 - 0.999899, this is the p-value you want to use for an upper tailed test.
Question 4 of 20
1.0/ 1.0 Points
A high school is running a campaign against the over-use of technology in teens. The committee
running the campaign decides to look at the difference in social media usage between teens and
adults. They take a random sample of 200 teens in their city (Group 1) and find that 85% of
them use social media, and then take another random sample of 180 adults in their city (Group 2)
and find that 55% of them use social media. Find a 90% confidence interval for the difference in
proportions.
Enter the confidence interval - round to 4 decimal places.
0.2262 < p1 - p2 < 0.3738
Answer Key:0.2262|.2262, 0.3738|.3738
Feedback:
Z-Critical Value =NORM.S.INV(.95) = 1.645
LL = (.85-.55) - 1.645*
UL = (.85-.55) + 1.645*
Question 5 of 20
0.5/ 1.0 Points
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.
0.356 < p1 - p2 < 0.139
Answer Key:-0.356|-.356, 0.139|.139
Feedback:
Z-Critical Value =NORM.S.INV(.995) = 2.575
LL = (.625-.7333) - 2.575*
UL = (.625-.7333) + 2.575*
Part 3 of 7 - Hypothesis Testing Questions 3.0/ 5.0 Points
Question 6 of
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