What is the test of significance for the coefficient of correlation. (Method 1 and Method 2). The second image is the guide. Thank you. I hope you can help me answer this.

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Author:Amos Gilat
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What is the test of significance for the coefficient of correlation. (Method 1 and Method 2). The second image is the guide. Thank you. I hope you can help me answer this.

INTERPRETATION OF COEFFICIENT OF CORRELATION
Strength of Correlation
Perfect positive correlation
Strong positive correlation
Moderately positive correlation
Weak positive correlation
Negligible positive correlation
Value of r
+1
+ 0.71 to + 0.99
+ 0.51 to + 0.70
+ 0.31 to + 0.50
+ 0.01 to + 0.30
No correlation
Negligible negative correlation
Weak negative correlation
Moderately negative correlation
Strong negative correlation
Perfect negative correlation
- 0.01 to - 0.30
- 0.31 to - 0.50
- 0.51 to - 0.70
- 0.71 to - 0.99
-1
b.) Does this degree of relationship is true for every year? (Use 0.05 level of significance)
TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 1)
1. Formulation of Null and Alternative Hypotheses
Null Hypothesis
Ho: p = 0: There is no significant relationship between the total enrolment and the yearly profit
Alternative Hypothesis
Ha: p#0: There is a significant relationship between the total enrolment and the yearly profit
2. Select the level of significance
a = 0.05
3. Test Statistics
МЕТНOD 1
6.
; t-test ; two
tailed test
n =
TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 1)
4. Define the Area of Rejection or the Critical Region
МЕТНOD 1:
dp: n-2
:1-1
-1 AU50
L. 8651
>
DECISION RULE:
Reject H, if the computed value is
a.) > 2.1657
b.)_< -1.9450
or
Otherwise, do not reject the Ho.
TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 1)
5. Compute for the
-value:
n- 2
0.43 x
> 1.907105 999 8.911)
7. Interpretation:
Rejection
there is a significant relationship between the
total enrolment and the yearly profit
6. Decision:
of the null hypothesis (Ho) means that
Since, the computed _t_-value ( 3.9371 ) is
greater than the tabular value ( 2.3560 ).
Therefore,
reject the null hypothesis
(Ho) at 0.05_level of significance.
base on the sample of
9 years
using
0.05
from 2011 to 2019
tev
>
level of significance.
Mejact
TEST OF SIGNIFICANCE FOR THE
COEFFICIENT OF CORRELATION
METHOD 2
b.) Does this degree of relationship is true for every year? (Use 0.05 level of significance)
TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 2)
1. Formulation of Null and Alternative Hypotheses
Null Hypothesis
Ho: p = 0: There is no significant relationship between the total enrolment and the yearly profit
Alternative Hypothesis
Ha: p#0: There is a significant relationship between the total enrolment and the yearly profit
2. Select the level of significance
a = 0.05
TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 2)
3. Test Statistics
МЕТНOD 2
n = _9_ ; L-test (df : n-k)
Reject
Fail to reject
p= 0
Reject
p = 0
0.89
-1 = -0.6660
crifral whne
K = 0.6660
Critical wolue
1
r =
0.83
TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 2)
4. Decision:
Since, the computed_-value (_
0.83
_) is _ greater than
the tabular
value ( 0.6660 ). Therefore,
_reject
the null hypothesis (Ho) at
_ 0.05 _ level of significance.
5. Interpretation:
Rejection
of
the
null
hypothesis
(Но)
means
that
there is a significant relationship between the total
enrolment and the yearly profit
9 years
from 2011 to 2019
base on the sample of
using
0.05
level of
significance.
Transcribed Image Text:INTERPRETATION OF COEFFICIENT OF CORRELATION Strength of Correlation Perfect positive correlation Strong positive correlation Moderately positive correlation Weak positive correlation Negligible positive correlation Value of r +1 + 0.71 to + 0.99 + 0.51 to + 0.70 + 0.31 to + 0.50 + 0.01 to + 0.30 No correlation Negligible negative correlation Weak negative correlation Moderately negative correlation Strong negative correlation Perfect negative correlation - 0.01 to - 0.30 - 0.31 to - 0.50 - 0.51 to - 0.70 - 0.71 to - 0.99 -1 b.) Does this degree of relationship is true for every year? (Use 0.05 level of significance) TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 1) 1. Formulation of Null and Alternative Hypotheses Null Hypothesis Ho: p = 0: There is no significant relationship between the total enrolment and the yearly profit Alternative Hypothesis Ha: p#0: There is a significant relationship between the total enrolment and the yearly profit 2. Select the level of significance a = 0.05 3. Test Statistics МЕТНOD 1 6. ; t-test ; two tailed test n = TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 1) 4. Define the Area of Rejection or the Critical Region МЕТНOD 1: dp: n-2 :1-1 -1 AU50 L. 8651 > DECISION RULE: Reject H, if the computed value is a.) > 2.1657 b.)_< -1.9450 or Otherwise, do not reject the Ho. TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 1) 5. Compute for the -value: n- 2 0.43 x > 1.907105 999 8.911) 7. Interpretation: Rejection there is a significant relationship between the total enrolment and the yearly profit 6. Decision: of the null hypothesis (Ho) means that Since, the computed _t_-value ( 3.9371 ) is greater than the tabular value ( 2.3560 ). Therefore, reject the null hypothesis (Ho) at 0.05_level of significance. base on the sample of 9 years using 0.05 from 2011 to 2019 tev > level of significance. Mejact TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION METHOD 2 b.) Does this degree of relationship is true for every year? (Use 0.05 level of significance) TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 2) 1. Formulation of Null and Alternative Hypotheses Null Hypothesis Ho: p = 0: There is no significant relationship between the total enrolment and the yearly profit Alternative Hypothesis Ha: p#0: There is a significant relationship between the total enrolment and the yearly profit 2. Select the level of significance a = 0.05 TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 2) 3. Test Statistics МЕТНOD 2 n = _9_ ; L-test (df : n-k) Reject Fail to reject p= 0 Reject p = 0 0.89 -1 = -0.6660 crifral whne K = 0.6660 Critical wolue 1 r = 0.83 TEST OF SIGNIFICANCE FOR THE COEFFICIENT OF CORRELATION (METHOD 2) 4. Decision: Since, the computed_-value (_ 0.83 _) is _ greater than the tabular value ( 0.6660 ). Therefore, _reject the null hypothesis (Ho) at _ 0.05 _ level of significance. 5. Interpretation: Rejection of the null hypothesis (Но) means that there is a significant relationship between the total enrolment and the yearly profit 9 years from 2011 to 2019 base on the sample of using 0.05 level of significance.
It is expected that students who consistently perform
well in major tests, quizzes, assignments and seatworks
in the classroom also perform well in standardized
achievement test. A teacher decided to examine this
hypothesis. At the end of the academic year, she
randomly selected 10 students from her class. She
computes a correlation between the student's
achievement test scores and the overall GPA for each
student computed over the entire year. The data for the
10 students from her class are provided below.
Student Achievement
GPA
A
98
1.2
95
1.3
C
90
1.6
D
87
1.9
E
62
3.5
a. Compute the correlation coefficient.
F
85
2.0
b. Interpret the result in A.
c. Determine the regression equation.
G
76
2.9
H
88
1.5
d. If a student scored a 90 on the achievement test,
what would be his predicted GPA?
71
3.2
e. What could be a students' predicted GPA if he/she
scored 75 and 85?
92
1.4
Transcribed Image Text:It is expected that students who consistently perform well in major tests, quizzes, assignments and seatworks in the classroom also perform well in standardized achievement test. A teacher decided to examine this hypothesis. At the end of the academic year, she randomly selected 10 students from her class. She computes a correlation between the student's achievement test scores and the overall GPA for each student computed over the entire year. The data for the 10 students from her class are provided below. Student Achievement GPA A 98 1.2 95 1.3 C 90 1.6 D 87 1.9 E 62 3.5 a. Compute the correlation coefficient. F 85 2.0 b. Interpret the result in A. c. Determine the regression equation. G 76 2.9 H 88 1.5 d. If a student scored a 90 on the achievement test, what would be his predicted GPA? 71 3.2 e. What could be a students' predicted GPA if he/she scored 75 and 85? 92 1.4
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