In: Statistics and Probability

If you wanted to calculate a 90% confidence interval for the
difference in average

number of friendship contacts between primary aged boys and girls
and we are

pretending that df=12, what t scores would you use? (assuming equal
variances again)

A. ☐+/- 1.356

B. ☐+/- 2.681

C. ☐+/- 1.782

D. ☐+/- 2.179

E. ☐+/- 3.055

9. Suppose you calculated your 90% interval as described above and
your lower

confidence limit was

–2.75 and your upper confidence limit was 3.20. What would that
mean?

A. ☐It would mean that boys have 2.75 fewer contacts than girls on
average

B. ☐It means that girls have 8.95 more contacts on average than
boys

C. ☐It means that there may be no difference between the average
number of

contacts for boys and girls

D. ☐It means that girls definitely have more contacts than
boys

E. ☐It means that girls have 3.20 times more friendship contacts
than boys

10. If you were to increase your confidence level to 99%, holding
everything else

constant

A. ☐Your interval would be more precise

B. ☐You would be less likely to miss the population value

C. ☐Your interval would be wider

D. ☐You would have less confidence

E. ☐You would have to change your df

Given

Confidence Level = 90%

df = 12 Degrees of
Freedom

For 90% Confidence interval

α = (100-90)/100 = 0.10

Degrees of freedom = df = 12

From the t-tables, or Excel function t.inv.2t(α,
df)

t = t.inv.2t(0.10, 12)
(We take the positive value for
calculations)

t = 1.782

Answer :

**C. ☐+/- 1.782
**

9) Confidence interval is (-2.75, 3.20)

Thus, 0 is part of the confidence interval

which implies that there is a possibility that

"there may be no difference between the average number
of

contacts for boys and girls"

Answer :

** C. ☐ It
means that there may be no difference between the average number
of
contacts for boys and girls **

10) Confidence interval is given by

When we increase the confidence level to 99%

the t-score will increase

Thus the value of

will increase

This will increase the width of the confidence interval

Answer :

** C. ☐ Your interval would be
wider **

You are interested in finding a 90% confidence interval for the
average number of days of class that college students miss each
year. The data below show the number of missed days for 14 randomly
selected college students. Round answers to 3 decimal places where
possible. 1 8 10 9 1 3 1 8 1 0 2 1 5 8 a. To compute the confidence
interval use a distribution. b. With 90% confidence the population
mean number of days of...

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 15 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible.
13
26
16
13
6
27
26
6
13
25
6
11
20
14
16
a. To compute the confidence interval use a ? z
t distribution.
b. With 90% confidence the population mean number of visits per
physical...

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 12 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible.
5
12
18
17
28
9
10
6
16
28
5
13
b. With 90% confidence the population mean number of visits per
physical therapy patient is between and
visits.
c. If many groups of 12 randomly selected physical...

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 13 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible. 19 19
14 13 26 19 5 14 17 27 13 20 28 a. To compute the confidence
interval use a distribution. b. With 90% confidence the population
mean number of visits per physical therapy patient is between...

You are interested in finding a 90% confidence interval for the
mean number of visits for physical therapy patients. The data below
show the number of visits for 12 randomly selected physical therapy
patients. Round answers to 3 decimal places where possible.
19
25
10
7
19
24
11
27
16
19
24
25
a. To compute the confidence interval use a ? z
t distribution.
b. With 90% confidence the population mean number of visits per
physical therapy patient is...

For this activity, you will be creating a confidence interval
for the average number of hours of TV watched. You will also run a
hypothesis test to test the claim that the average number of hours
of TV watched is more than 7 hours. Students from a former MAT 152
class asked the question, “How many hours of TV do you watch in a
typical week?” Please use the data they collected to answer the
following questions.
Data: 15, 11,...

For this activity, you will be creating a confidence interval
for the average number of hours of TV watched. Last semester, my
MAT 152 online students asked the question, “How many hours of
screen time do you have in a typical week?”
Please use the data they collected to answer the following
questions. Data: 21, 2, 28, 30, 18, 21, 25, 20, 25, 14, 21, 25, 50,
39, 46, 20, 35, 45, 37, 46
Name at least 2 ways this...

For this activity, you will be creating a confidence interval
for the average number of hours of TV watched. Last semester, my
MAT 152 online students asked the question, “How many hours of
screen time do you have in a typical week?” Please use the data
they collected to answer the following questions. Data: 21, 2, 28,
30, 18, 21, 25, 20, 25, 14, 21, 25, 50, 39, 46, 20, 35, 45, 37, 46
Are there any outliers in this...

A newsgroup is interested in constructing a 90% confidence
interval for the difference in the proportions of Texans and New
Yorkers who favor a new Green initiative. Of the 506 randomly
selected Texans surveyed, 436 were in favor of the initiative and
of the 525 randomly selected New Yorkers surveyed, 459 were in
favor of the initiative.
a. With 90% confidence the difference in the proportions of Texans
and New Yorkers who favor a new Green initiative is between _______...

A newsgroup is interested in constructing a 90% confidence
interval for the difference in the proportions of Texans and New
Yorkers who favor a new Green initiative. Of the 582 randomly
selected Texans surveyed, 428 were in favor of the initiative and
of the 569 randomly selected New Yorkers surveyed, 456 were in
favor of the initiative.
. With 90% confidence the difference in the proportions of
Texans and New Yorkers who favor a new Green initiative is between
(round...

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