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xenn [34]
1 year ago
13

A law school administrator was interested in whether a student's score on the entrance exam can be used to predict a student's g

rade point average (GPA) after one year of law school. The administrator studied 15 students. It was shown that the correlation between the entrance exam score and the grade point average after one year of law school was 0.934. Based on this information, interpret the correlation coefficient.
Mathematics
1 answer:
frutty [35]1 year ago
8 0

Answer:

The correlation coeffcient for this case was provided:

r =0.934

And this coefficient is very near to 1 the maximum possible value, so then we can interpret that the relationship between the entrace exam score and the grade point average are strongly linearly correlated .

We can also find the r^2 who represent the determination coefficient and we got:

r^2 = 0.934^2= 0.872

And the interpretation for this is that a linear model explains appproximately 87.2% of the variability between the two variables

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

The correlation coeffcient for this case was provided:

r =0.934

And this coefficient is very near to 1 the maximum possible value, so then we can interpret that the relationship between the entrace exam score and the grade point average are strongly linearly correlated .

We can also find the r^2 who represent the determination coefficient and we got:

r^2 = 0.934^2= 0.872

And the interpretation for this is that a linear model explains appproximately 87.2% of the variability between the two variables

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p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

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t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

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The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

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If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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