Which scatterplot has a correlation coefficient closest to r= 1?
2 answers:
You might be interested in
Answer:
a. Line
b. Plane
c. All of R^3
Step-by-step explanation:
In order to answer this question, we need to study the linear independence between the vectors :
1 - A set of three linearly independent vectors in R^3 generates R^3.
2 - A set of two linearly independent vectors in R^3 generates a plane.
3 - A set of one vector in R^3 generates a line.
The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :
a. Let's put the vectors into the rows of a matrix :
⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).
We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).
At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.
b. Again, let's put the vectors into the rows of a matrix :
⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).
c. Finally :
⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒
The set is linearly independent so the set of all linear combination of the set c. is all of R^3.
<h2>Answer with explanation:</h2>
Let be the average starting salary ( in dollars).
As per given , we have
Since is left-tailed , so our test is a left-tailed test.
WE assume that the starting salary follows normal distribution .
Since population standard deviation is unknown and sample size is small so we use t-test.
Test statistic : , where n= sample size ,
= sample mean , s = sample standard deviation.
Here , n= 15 , , s= 225
Then,
Degree of freedom = n-1=14
The critical t-value for significance level α = 0.01 and degree of freedom 14 is 2.62.
Decision : Since the absolute calculated t-value (2.07) is less than the critical t-value., so we cannot reject the null hypothesis.
Conclusion : We do not have sufficient evidence at 1 % level of significance to support the claim that the average starting salary of the graduates is significantly less that $42,000.