Access Millions of academic & study documents

Chi

Content type
User Generated
Subject
Statistics
School
nua
Type
Homework
Showing Page:
1/7
Question 1:
Right Ear
Left Ear
No preference
Total
Right-handed
446
172
42
660
Left-handed
12
47
04
63
Total
458
219
46
723
As we know, Expected Frequency
( ) ( )
Total Grand
Total ColumnTota Row
=
(a) Expected Value for the cell that has an observed frequency of 4
008.4
723
6346
=
=
(b) Because the expected frequency of a cell is less than 5, the requirements for the
hypothesis test are not satisfied.
Question 2:
Test statistic is 68.738
The p-value is 0.000
(1) Less than (2) Reject (3) is
Question 3:
Determine the Null and the alternate hypothesis
Option (A) is answer.
Observed Frequency:
Expected Frequency:

Sign up to view the full document!

lock_open Sign Up
Showing Page:
2/7
Test statistic
( )
532.759.362.276.056.0
2
1
2
1
2
2
2
=+++=
=
= =i j
ij
ijij
E
EO
P-value
( )
006.0532.7
2
=
P
(1) less than (2) Reject (3) is (4) surgery treatment is better.
Question 4:
Determine the Null and the alternate hypothesis
Option (B) is answer.
Observed Frequency:
Expected Frequency:
Test statistic
( )
071.382.079.075.072.0
2
1
2
1
2
2
2
=+++=
=
= =i j
ij
ijij
E
EO
P-value
( )
080.0071.3
2
=
P
(1) Greater than (2) Fail to reject (3) is not (4) do not appear

Sign up to view the full document!

lock_open Sign Up
Showing Page:
3/7

Sign up to view the full document!

lock_open Sign Up
End of Preview - Want to read all 7 pages?
Access Now
Unformatted Attachment Preview
Question 1: Right-handed Left-handed Total Right Ear 446 12 458 Left Ear 172 47 219 As we know, Expected Frequency = No preference 42 04 46 (Row Tota )  (Column Total 660 63 723 Total ) Grand Total 46  63 = 4.008 723 (b) Because the expected frequency of a cell is less than 5, the requirements for the hypothesis test are not satisfied. (a) Expected Value for the cell that has an observed frequency of 4 = Question 2: Test statistic is 68.738 The p-value is 0.000 (1) Less than (2) Reject (3) is Question 3: Determine the Null and the alternate hypothesis Option (A) is answer. Observed Frequency: Splint Treatment Surgery Treatment Successful Treatment 67 60 Unsuccessful Treatment 22 05 Successful Treatment 73.40 53.60 Unsuccessful Treatment 15.60 11.40 Expected Frequency: Splint Treatment Surgery Treatment Test statistic 2 2  =  2 (O − Eij ) 2 ij = 0.56 + 0.76 + 2.62 + 3.59 = 7.532 E 2 ij i =1 j =1 P-value ( ) P  2  7.532 = 0.006 (1) less than (2) Reject (3) is (4) surgery treatment is better. Question 4: Determine the Null and the alternate hypothesis Option (B) is answer. Observed Frequency: Adverse Health condition reported No Adverse Health condition reported Amalgum 132 Composite 147 140 115 Expected Frequency: Adverse Health condition reported No Adverse Health condition reported Amalgam 142.11 Composite 136.89 129.89 125.11 Test statistic 2 2  =  2 (O ij − Eij ) 2 E 2 ij i =1 j =1 = ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
Studypool
4.7
Indeed
4.5
Sitejabber
4.4