In: Statistics and Probability
Claim: There is a significant correlation between test 1 scores and the hours a student spent studying.
Use the data-analysis add-in to complete the hypothesis test for the correlation coefficient using level of significance of 0.05.
Hours Spent Studying | Test 1 Scores |
8 | 91 |
6 | 83 |
6 | 75 |
12 | 85 |
4 | 90 |
6 | 80 |
7 | 67 |
7 | 82 |
5 | 88 |
9 | 87 |
5 | 95 |
3 | 91 |
2 | 73 |
13 | 80 |
4 | 83 |
10 | 92 |
12 | 94 |
0 | 68 |
3 | 75 |
8 | 91 |
3 | 79 |
11 | 95 |
0 | 38 |
8 | 76 |
9 | 91 |
7 | 85 |
0 | 59 |
2 | 70 |
1 | 69 |
5 | 78 |
2. Hypothesis Test | identify the claim | ||
Hypothesis Statements | Ho | ||
Ha | |||
Standardized Test Statistic | |||
P-value | |||
Insert Data Analysis test here |
First we calculate the correlation coefficient between test 1 scores and the hours a student spent studying:
X | Y | X*Y | X2 | Y2 | |
8 | 8 | 64 | 64 | 64 | |
6 | 6 | 36 | 36 | 36 | |
6 | 6 | 36 | 36 | 36 | |
12 | 12 | 144 | 144 | 144 | |
4 | 4 | 16 | 16 | 16 | |
6 | 6 | 36 | 36 | 36 | |
7 | 7 | 49 | 49 | 49 | |
7 | 7 | 49 | 49 | 49 | |
5 | 5 | 25 | 25 | 25 | |
9 | 9 | 81 | 81 | 81 | |
5 | 5 | 25 | 25 | 25 | |
3 | 3 | 9 | 9 | 9 | |
2 | 2 | 4 | 4 | 4 | |
13 | 13 | 169 | 169 | 169 | |
4 | 4 | 16 | 16 | 16 | |
10 | 10 | 100 | 100 | 100 | |
12 | 12 | 144 | 144 | 144 | |
0 | 0 | 0 | 0 | 0 | |
3 | 3 | 9 | 9 | 9 | |
8 | 8 | 64 | 64 | 64 | |
3 | 3 | 9 | 9 | 9 | |
11 | 11 | 121 | 121 | 121 | |
0 | 0 | 0 | 0 | 0 | |
8 | 8 | 64 | 64 | 64 | |
9 | 9 | 81 | 81 | 81 | |
7 | 7 | 49 | 49 | 49 | |
0 | 0 | 0 | 0 | 0 | |
2 | 2 | 4 | 4 | 4 | |
1 | 1 | 1 | 1 | 1 | |
5 | 5 | 25 | 25 | 25 | |
Sum = | 176 | 176 | 1430 | 1430 | 1430 |
The correlation coefficient r is computed using the following expression:
where
In this case, based on the data provided, we get that
Therefore, based on this information, the sample correlation coefficient is computed as follows
The following needs to be tested:
H0:ρ=0
HA:ρ̸0
where \rhoρ corresponds to the population correlation.
The sample size is n=30, so then the number of degrees of freedom is df=n−2=30−2=28
The corresponding t-statistic to test for the significance of the correlation is:
The p-value is computed as follows:
p=Pr(∣t28∣>)<0.0001
Since we have that p<0.0001, it is concluded that the null hypothesis H0 is rejected.
Therefore, based on the sample correlation provided, it is concluded that there is enough evidence to claim that the population correlation ρ is different than 0, at the 0.05 significance level.
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