Calculating the Pearson correlation with x -scores Imagine that a professor of a

Question

Calculating the Pearson correlation with x -scores Imagine that a professor of anthropology has two teaching assistants (TAs) who will help her grade assignments for the duration of the semester. The professor wants to make that she and the TAs are well calibrated with one another, so she has all three of them grade the first assignment independently. Because the professor grades every assignment on a curve, she first converts the students' scores to x-scores for each grader. The flowing table shows the x-scores for population of 10 students in her class for each grader. The professor is going to use the z-scores to calculate the Pearson correlation coefficient between her scores and those of her TAs. To calculate the Pearson correlation coefficient, she sums the products of the x-scores. The professor should divide this sum by ____ The professor constructed a table of correlation coefficients between her scores and those of her TAs. Select the correct missing correlation. B0ased on the correlation table, the professor should arrive at which of the following conclusions? The professor and TA #1 are well calibrated, but TA # 2 is off. The professor and both of her TAs are well calibrated. The professor and TA #2 are well calibrated, but TA #1 is off. The two TAs ore well calibrated with each other but not with the professor. The professor thinks she tends to be a harsh grader and decides to multiply the grades she gave each student by 1.05 before converting the scores to z-scores and computing the Pearson correlation coefficient between her grades and the two teaching assistants. This change in the professor's scores will ____the correlation between the professor's scores and TA #1's and TA #2's scores.

Explanation / Answer

The professor should divide this sum by sum of squares of the z-scores.
Enter the data into Excel and carry out REGRESSION from DATA ANALYSIS TOOLPACK.

Professor

1

Teaching Assistant #1

0.238901

1

Teaching Assistant #2

0.360716

0.899865

The data is entered into Excel and saved in CSV format.
R code :
data1=read.csv(file.choose(),header=T)
attach(data1)
cor.test(Professor,X1)

Output :

Pearson's product-moment correlation

data: Professor and X1

t = 0.69586, df = 8, p-value = 0.5062

alternative hypothesis: true correlation is not equal to 0

95 percent confidence interval:

-0.4599032 0.7549665

sample estimates:

     cor

0.238901

cor.test(Professor,X2)

Output :

Pearson's product-moment correlation

data: Professor and X2

t = 1.0939, df = 8, p-value = 0.3058

alternative hypothesis: true correlation is not equal to 0

95 percent confidence interval:

-0.3479313 0.8070484

sample estimates:

      cor

0.3607158

cor.test(X1,X2)

Output :

Pearson's product-moment correlation

data: X1 and X2

t = 5.8354, df = 8, p-value = 0.0003892

alternative hypothesis: true correlation is not equal to 0

95 percent confidence interval:

0.6235009 0.9763259

sample estimates:

      cor

0.8998649

If p-value is > 0.05, we accept the null hypothesis of zero correlation at 5% level.
Thus, from the p-values we get,
The professor and TA#2 are well calibrated and TA#1 is off.

Professor

1

Teaching Assistant #1

0.238901

1

Teaching Assistant #2

0.360716

0.899865

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