Indiana State University Statistics Hypothesis Testing Questions

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Hello I need correct answers for 2 statistics questions ASAP. Question 1 and question 2, there are sub questions for both questions and please answer all of them. I have attached pictures of these two questions. Thank you.

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Test
Null hypothesis He: Hi-Ma = 0
Alternative hypothesis HV – H2 > 0
T-Value DF P-value
0.12 593 0.453
Output 3
664
Test
Null hypothesis Hoipl. – Hz = 0
Alternative hypothesis Hipo-Pa = 0
T-Value DF P-Value
0.12 593 0.905
d. Which of the above outputs is the correct Minitab output for this test.
[ Select]
e. If you did not have computer software available, what would the conservative degrees of freedom
have been for this significance test? [Select ]
f. If the assumption of normality had not been met for this significance test, what other method
could have been used to compare the two means? (Select]
us
Next
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A researcher is interested to see if there is significant evidence that Freshmen sleep more than
Others.
Possible Minitab outputs for this significance test are given below.
Output 1
Test
Null hypothesis Hoi Ho – Mz = 0
Alternative hypothesis Hip-H: < 0 T-Value DF P-value 0.12 593 0.547 Output 2 Test Null hypothesis Hai Mi-P: = 0 Altemative hypothesis Hu-Pz > 0
T-Value DF P-value
0.12 593 0.453
Output 3
Test
Null hypothesis Ho: Mo – Mz = 0
Alternative hypothesis Hiiu-M: 0
T-Value DF P-value
0.12 593 0.905
ho
d. Which of the above outputs is the correct Minitab output for this test.
[ Select]
e. If you did not have computer software avai || app.honorlock,com is sharing your screen.
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d. The p-value is 0.000, what conclusion can you make?
Fill in the blank.
With a p-value of 0.000 (small p-value),
There is
evidence that the true
proportion of Americans that get most of their information about current events from the Internet is
greater in 2018 than in 2012. [Select ]
23
A 95% confidence interval was calculated using the same set of data. The Minitab output is given
below.
Descriptive Statistics
Sample N
Event Sample p
Sample 1 994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
95% CI for
Difference Difference
-0.158309 (-0.198625, -0.117993)
e. Complete the interpretation.
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 is between
than in 2018.
[ Select]
f. Given the following two statements:
1. The necessary assumptions were met for both the significance test and the 95% confidence
interval.
II. In the Minitab output above, Difference
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Attempt due: Apr 24 at 3:10pm
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COOTER
Has the way in which people get information about current events in the news changed in recent
times? A General Social Survey found that in 2012 (group 1), 299 out of 994 Americans got most of
their information about current events from the Internet. In 2018 (group 2), 539 out of 1174
Americans got most of their information about current events from the Internet. Is there significant
evidence that the true proportion of Americans that get most of their information about current
events from the Internet is greater in 2018 than in 2012?
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664
Live Chat
a. Is this a question about independent or dependent proportions? [Select]
Phone
b. What is the correct hypothesis for this significance test?
Note: In the answer choices “not =” stands for +
[ Select]
c. The calculation for the test statistic is:
2 =
(0.3008-0.4591)-0
10.3865(1-0.3865) + 117)
= -7.54
Which of these values is the pooled proportion? ! [ Select]
d. The p-value is 0.000, what conclusion can you make?
Fill in the blank.
With a p-value of 0.000 (small p-value), I
There is
evidence that the true
proportion of Americans that get most of their information about current events from the Internet is
greater in 2018 than in 2012. [Select]
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B
A
B
N
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below.
Descriptive Statistics
Sample N
Event Sample p
Sample 1 994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
3
Difference
-0.158309
95% CI for
Difference
(-0.198625, -0.117993)
e. Complete the interpretation.
?
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 between
than in 2018.
[ Select]
f. Given the following two statements:
w
1. The necessary assumptions were met for both the significance test and the 95% confidence
interval.
11. In the Minitab output above, Difference = -0.158309 represents the point estimate P1 – P2.
Which of the statements are true? [Select]
V
< Previous Next → Il app.honorlock.com is sharing your screen. Stop sharing Hide Ruiz Honorlock Browser Guard™ 国家发国看到的。虽時本着车到车 , -O0OZGAT LOVEADSY 4X2 vu, 10 I N M Question 1 SUSCRON 18 pts Question 8 6 OGO Announcements Honorlock Time Running: Hide Attempt due: Apr 24 at 3:10pm 1 Hour, 39 Minutes, 27 Seconds GatorEvals How do students in their first year of college differ from students who have been in college longer? A recent student survey asked a random sample of 337 freshmen (group 1) and 286 others (group 2) how many hours they sleep on a typical night. Data for freshmen and others is given in the table below. 36 Zoom Conferences 24/7 Proctoring Support Freshmen (group 1) Others (group 2) Calculator Sample Mean, I 7.558 7.547 Live Chat Sample Standard Deviation, s 1.080 1.137 Sample Size, n 0 Phone 337 286 3 a. Which of the following are the necessary assumptions for the 95% confidence interval for comparing two independent means? 1. Quantitative data in both groups II. Categorical data in both groups III. Independent random samples IV. The sample sizes n1 = 337 and n2 = 286 are both at least 30 V. Successes in group 1 = 337 and successes in group 2 = 286, both are at least 15 For credit, you must choose all of the assumptions that apply. [Select] b. What is the parameter? [ Select] / C. Calculate the 95% confidence interval. [Select ] Il app.honorlock.com is sharing your screen. Stop sharing Hide Clonorlock Browser Guard™ 享 DOZENT LOVENDSY V W 8 O) R o S G H K N M M al A A researcher is interested to see if there is significant evidence that Freshmen sleep more than Others. Possible Minitab outputs for this significance test are given below. Output 1 Test Null hypothesis Hoi Ho - Mz = 0 Alternative hypothesis Hip-H: < 0 T-Value DF P-value 0.12 593 0.547 Output 2 Test Null hypothesis Hai Mi-P: = 0 Altemative hypothesis Hu-Pz > 0
T-Value DF P-value
0.12 593 0.453
Output 3
Test
Null hypothesis Ho: Mo – Mz = 0
Alternative hypothesis Hiiu-M: 0
T-Value DF P-value
0.12 593 0.905
ho
d. Which of the above outputs is the correct Minitab output for this test.
[ Select]
e. If you did not have computer software avai || app.honorlock,com is sharing your screen.
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below.
Descriptive Statistics
Sample N
Event Sample p
Sample 1 994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
3
Difference
-0.158309
95% CI for
Difference
(-0.198625, -0.117993)
e. Complete the interpretation.
?
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 between
than in 2018.
[ Select]
f. Given the following two statements:
w
1. The necessary assumptions were met for both the significance test and the 95% confidence
interval.
11. In the Minitab output above, Difference = -0.158309 represents the point estimate P1 – P2.
Which of the statements are true? [Select]
V
< Previous Next → Il app.honorlock.com is sharing your screen. Stop sharing Hide Ruiz Honorlock Browser Guard™ 国家发国看到的。虽時本着车到车 , -O0OZGAT LOVEADSY 4X2 vu, 10 I N M d. The p-value is 0.000, what conclusion can you make? Fill in the blank. With a p-value of 0.000 (small p-value), There is evidence that the true proportion of Americans that get most of their information about current events from the Internet is greater in 2018 than in 2012. [Select ] 23 A 95% confidence interval was calculated using the same set of data. The Minitab output is given below. Descriptive Statistics Sample N Event Sample p Sample 1 994 299 0.300805 Sample 2 1174 539 0.459114 Estimation for Difference 95% CI for Difference Difference -0.158309 (-0.198625, -0.117993) e. Complete the interpretation. We are 95% confident that the true proportion of Americans that get most of their information about current events from the Internet in 2012 is between than in 2018. [ Select] f. Given the following two statements: 1. The necessary assumptions were met for both the significance test and the 95% confidence interval. II. In the Minitab output above, Difference Il app.honorlock.com is sharing your screen. Stop sharing Hide Honorlock Browser Guard™ DEZIBEL *14. 000Z LOVENJOYEN - 66 0 o Z 11 N M Question 1 SUSCRON 18 pts Question 8 6 OGO Announcements Honorlock Time Running: Hide Attempt due: Apr 24 at 3:10pm 1 Hour, 39 Minutes, 27 Seconds GatorEvals How do students in their first year of college differ from students who have been in college longer? A recent student survey asked a random sample of 337 freshmen (group 1) and 286 others (group 2) how many hours they sleep on a typical night. Data for freshmen and others is given in the table below. 36 Zoom Conferences 24/7 Proctoring Support Freshmen (group 1) Others (group 2) Calculator Sample Mean, I 7.558 7.547 Live Chat Sample Standard Deviation, s 1.080 1.137 Sample Size, n 0 Phone 337 286 3 a. Which of the following are the necessary assumptions for the 95% confidence interval for comparing two independent means? 1. Quantitative data in both groups II. Categorical data in both groups III. Independent random samples IV. The sample sizes n1 = 337 and n2 = 286 are both at least 30 V. Successes in group 1 = 337 and successes in group 2 = 286, both are at least 15 For credit, you must choose all of the assumptions that apply. [Select] b. What is the parameter? [ Select] / C. Calculate the 95% confidence interval. [Select ] Il app.honorlock.com is sharing your screen. Stop sharing Hide Clonorlock Browser Guard™ 享 DOZENT LOVENDSY V W 8 O) R o S G H K N M M al A IV. The sample sizes ni = 337 and n2 = 286 are both at least 30 V. Successes in group 1 = 337 and successes in group 2 = 286, both are at least 15 For credit, you must choose all of the assumptions that apply. [Select] b. What is the parameter? Bu [ Select] c. Calculate the 95% confidence interval. (Select] < ? (Select) (7.443, 7.763) men sleep more than (-0.078, 0.100) A researcher is interested to see if there is (7.415, 7.679) Others. (-0.164, 0.186) Possible Minitab outputs for this significance test are given below. Output 1 Test Null hypothesis Ho Ho - P2 = 0 Alternative hypothesis Hs: My - P: 0 T-Value DF P-Value 0.12 593 0.453 Il app.honorlock.com is sharing your screen. Stop sharing Hide uெtnut 2 TH sonorlock Browser Guard™ 等,希子等; . *DDOTECTO LOVEA394 VIEW 5 8 KOOS (3 ) E o A B N M Sampionida DeVianon, 11.VOU Sample Size, n 337 i Phone 286 co a. Which of the following are the necessary assumptions for the 95% confidence interval for comparing two independent means? 83 1. Quantitative data in both groups II. Categorical data in both groups III. Independent random samples IV. The sample sizes n1 = 337 and n2 = 286 are both at least 30 V. Successes in group 1 = 337 and successes in group 2 = 286, both are at least 15 66 For credit, you must choose all of the assumptions that apply. [ Select ] b. What is the parameter? [ Select ] [ Select ] xbar_1 - xbar_2, the difference in the sample mean hours of sleep on a typical night between freshmen and others. p_1-p_2, the difference in the true proportion of freshmen and others who sleep at night. mu_1 - mu_2, the difference in the true mean hours of sleep on a typical night between freshmen and others. mu_d, the true mean difference in hours of sleep on a typical night between freshmen and others. mu_1 - mu_2, the difference in the sample mean hours of sleep on a typical night between freshmen and others. A researcher is interested to see if there is significant evidence that Freshmen sleep more than Others. Possible Minitab outputs for this significance test are given below. Output 1 Test Null hypothesis Ho Ho - Hz = 0 Alternative hypothesis H: My - : < 0 T-Value DF P-Value 0.12 593 0547 Il app.honorlock.com is sharing your screen. Stop sharing Hide Slonorlock Browser Guard™ 多而水等等, -DOOZZOTT LOVEA194 VIEN = 5 9 o o a A K B N M Test Null hypothesis Hello : = 0 Alternative hypothesis He: ui - Mz > 0
T-Value DF P-value
0.12 593 0.453
Output 3
88
RODI
Dea
664
Test
Null hypothesis Hoi Ho – Me = 0
Alternative hypothesis HiiH – Pa = 0
T-Value DF P-value
0.12 593 0.905
(?
d. Which of the above outputs is the correct Minitab output for this test.
[ Select]
e. If you did not have computer software available, what would the conservative degrees of freedom
have been for this significance test? [
[ Select ]
f. If the assumption of normality had not been met for this significance test, what other method
could have been used to compare the two means? (elect]
[ Select ]
One sample t test for the mean
Permutation Test
Two sample z test for comparing independent proportions
McNemar’s Test
S
IIZ
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Output 2
Test
Null hypothesis Ho: Mi – M: = 0
Alterative hypothesis He: M – N2 > 0
T-Value DE P-value
0.12 593 0.453
Output 3
Test
Null hypothesis Ho Ho – P: = 0
Alternative hypothesis HP, – M27 0
T-Value
0.12
DF P-value
593 0.905
?
d. Which of the above outputs is the correct Minitab output for this test.
[ Select]
e. If you did not have computer software available, what would the conservative degrees of freedom
have been for this significance test? [Select ]
f. If the assumption of normality had not been met for this significance test, what other method
could have been used to compare the two means? [Select]
Next
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For credit, you must choose all of the assumptions that apply. (Select]
b. What is the parameter?
[ Select]
50
c. Calculate the 95% confidence interval. [Select]
664
A researcher is interested to see if there is significant evidence that Freshmen sleep more than
Others.
Possible Minitab outputs for this significance test are given below.
Output 1
Test
Null hypothesis Hoilla – Mz = 0
Alternative hypothesis Hy: M. – H2 < 0 T-Value 0.12 DF P-value 0.547 593 Output 2 Test Null hypothesis Ho: Hi - H: = 0 Alternative hypothesis Hai Mo - M2 > 0
T-Value
DF
P-Value
0.12
593
0.453
Output 3
Test
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and
Question 1
Question 7
Question 8
18 pts
ma
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Attempt due Apr 24 at 3:10pm
1 Hour, 5 Minutes. 40 Seconds
Hook
How do students in their first year of college differ from students who have been in college longer?
A recent student survey asked a random sample of 337 freshmen (group 1) and 286 others (group 2)
how many hours they sleep on a typical night. Data for freshmen and others is given in the table
below.
storal
38
24/7 Proctoring Support
Zoom Conferences
Freshmen (group 1)
Others (group 2)
E Calculator
Sample Mean,
664
7.558
7.547
77 Live Chat
Sample Standard Deviation, s
1.080
1.137
Phone
Sample Size, n
337
286
?
a. Which of the following are the necessary assumptions for the 95% confidence interval for
comparing two independent means?
1. Quantitative data in both groups
II. Categorical data in both groups
III. Independent random samples
IV. The sample sizes n1 = 337 and n2 = 286 are both at least 30
V. Successes in group 1 = 337 and successes in group 2 = 286, both are at least 15
For credit, you must choose all of the assumptions that apply. [Select]
the
b. What is the parameter?
[ Select]
c. Calculate the 95% confidence interval. [Select]
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d. The p-value is 0.000, what conclusion can you make?
Fill in the blank.
With a p-value of 0.000 (small p-value),
There is
evidence that the true
proportion of Americans that get most of their information about current events from the Internet is
greater in 2018 than in 2012. [Select ]
A 95% confidence interval was calculated using the same set of data. The Minitab output is given
below.
Descriptive Statistics
Sample N
Event Sample p
Sample 1 994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
95% CI for
Difference Difference
-0.158309 (-0.198625, -0.117993)
e. Complete the interpretation.
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 is between
than in 2018.
[ Select]
f. Given the following two statements:
tort and the OC0/montar
1. The necessary assumptions were met for hath tha sinnifer
interval.
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Question 2
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Question 8
18 pts
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ck
Has the way in which people get information about current events in the news changed in recent
times? A General Social Survey found that in 2012 (group 1), 299 out of 994 Americans got most of
their information about current events from the Internet. In 2018 (group 2), 539 out of 1174
Americans got most of their information about current events from the Internet. Is there significant
evidence that the true proportion of Americans that get most of their information about current
events from the Internet is greater in 2018 than in 2012?
24/7 Proctoring Support
Contences
Calculator
5 Live Chat
Phone
a. Is this a question about independent or dependent proportions? [Select]
b. What is the correct hypothesis for this significance test?
Note: In the answer choices “not =” stands for +
[ Select]
c. The calculation for the test statistic is:
2=
(0.3008-0.4591)-0
0.3865(1–0.3865)[;
–
-7.54
1
994
+ 117)
Which of these values is the pooled proportion? [Select] W
[ Select ]
0.3008
d. The p-value is 0.000, what conclusion can you
0.3865
Fill in the blank.
1/1174
0.4591
With a p-value of 0.000 (small p-value), L
proportion of Americans that get most of their in
ence that the true
from the Internet is
1/994
greater in 2018 than in 2012. [Select ]
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Question 8
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18 pts
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Has the way in which people get information about current events in the news changed in recent
times? A General Social Survey found that in 2012 (group 1), 299 out of 994 Americans got most of
their information about current events from the Internet. In 2018 (group 2), 539 out of 1174
Americans got most of their information about current events from the Internet. Is there significant
evidence that the true proportion of Americans that get most of their information about current
events from the Internet is greater in 2018 than in 2012?
24/7 Proctoring Support
RE
Zoom Conferences
Calculator
7 Live Chat
iPhone
a. Is this a question about independent or dependent proportions? [Select]
b. What is the correct hypothesis for this significance test?
Note: In the answer choices “not =” stands for #
[ Select ]
[ Select ]
Ho: phat_1 – phat_2 = 0 versus Ha: phat_1 – phat_2 not = 0
Ho: p_1-p_2 < 0 versus Ha: p_1-p_2 = 0 Ho: phat_1 - phat_2 0 Ho: p_1-p_2 = 0 versus Ha: p_1-p_2 < 0 Ho: p_1-p_2 = 0 versus Ha: p_1-p_2 > 0
ect]
d. The p-value is 0.000, what conclusion can you make?
Fill in the blank.
With a p-value of 0.000 (small p-value),
There is
evidence that the true
proportion of Americans that get most of their information about current events from the Internet is
greater in 2018 than in 2012. [Select ]
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Question 8
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18 pts
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Is
Has the way in which people get information about current events in the news changed in recent
times? A General Social Survey found that in 2012 (group 1). 299 out of 994 Americans got most of
their information about current events from the Internet. In 2018 (group 2), 539 out of 1174
Americans got most of their information about current events from the Internet. Is there significant
evidence that the true proportion of Americans that get most of their information about current
events from the Internet is greater in 2018 than in 2012?
24/7 Proctoring Support
3
or ferences
* Calculator
Live Chat
a. Is this a question about independent or dependent proportions? (Select]
Phone
b. What is the correct hypothesis for this significance test?
Note: In the answer choices “not=”
=” stan
fort
[ Select)
C. The calculation for the test statistic is:
z =
(0.3008–0.4591)-0
V0.3865(1–0.3865) 004 + 1176
= -7.54
Which of these values is the pooled proportion? [Select]
>
d. The p-value is 0.000, what conclusion can you make?
Fill in the blank.
With a p-value of 0.000 (small p-value),
There is
evidence that the true
proportion of Americans that get most of their information about current events from the Internet is
greater in 2018 than in 2012.
[ Select ]
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below.
Descriptive Statistics
Sample N
Event Sample p
Sample 1 994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
95% Cl for
Difference Difference
-0.158309 (-0.198625, -0.117993)
e. Complete the interpretation.
?
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 is between
than in 2018.
[ Select]
f. Given the following two statements:
1. The necessary assumptions were met for both the significance test and the 95% confidence
interval.
II. In the Minitab output above, Difference = -0.158309 represents the point estimate P1 – P2-
Which of the statements are true? [Select]
< Previous Next >
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A 95% conhdence interval was calculated using the same set of data. The Minitab output is given
below.
Descriptive Statistics
Sample N Event Sample
Sample 1
994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
95% CI for
Difference Difference
-0.158309 (-0.198625, -0.117993)
e. Complete the interpretation.
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 is between
than in 2018.
[ Select]
( Select]
0.199 more to 0.118 less
ments:
0.118 and 0.199 less
0.199 less to 0.118 more
were met for both the significance test and the 95% confidence
0.118 and 0.199 more
, Difference = -0.158309 represents the point estimate P1 – P2.
Which of the statements are true? [Select]
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Which of these values is the pooled proportion? [Select]
d. The p-value is 0.000, what conclusion can you make?
Fill in the blank.
33
With a p-value of 0.000 (small p-value),
There is
evidence that the true
proportion of Americans that get most of their information about current events from the Internet is
BRO
30)
664
greater in 2018 than in 2012. [Select ]
[ Select ]
fail to reject Ho; not enough
?
accept Ho; very strong
A 95% confidence interval was
below.
reject Ho; not enough
of data. The Minitab output is given
fail to reject Ha; some
Descriptive Statistics reject Ho; very strong
Sample N Event Sample p
Sample 1 994 299 0.300805
Sample 2 1174 539 0.459114
Estimation for Difference
95% CI for
Difference Difference
-0.158309 (-0.198625, -0.117993)
e. Complete the interpretation.
We are 95% confident that the true proportion of Americans that get most of their information
about current events from the Internet in 2012 is between
than in 2018.
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Tags:
statistics

null hypothesis

Binomial Distribution

random variables

algebraic operations

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