If the correlation was 0.6, perform a calculation that justifies your answer to

Question

If the correlation was 0.6, perform a calculation that justifies your answer to part a) The GPA the first year of college is a 2.1, with an SD of 0.8 while, the average GPA during the second year is 2.5, with an SD of 0.8. The correlation is 0.4, and the scatter plot is football shaped. Predict the second year GPA of someone who had a first year GPA of 2.5. Predict the first year GPA of someone who had a second year GPA of 3.0 Predict the second year GPA of someone who had a first year GPA of 2.1. Predict the first year GPA of someone with an unknown second year GPA. True or false: people tend to have a higher GPA their second year. Circle the best choice: Someone in the 40th percentile their first year tends to be True or False: The person at the top of the class the first year is predicted to be at the top of the class the second year.

Explanation / Answer

with correlation, r=0.4

a) Predicting second year GPA of someone who has the first year GPA of 2.5

Standardised value of first year is Z(first year GPA)=(X-Mean)/S.D.

=(2.5-2.1)/0.8

=0.4/0.8

=0.5 S.D.s away from mean

Hence, predicting standardized first year GPA, Z(Second year GPA)=correalation* Z(First year GPA)

Z(Second year GPA)=0.4*0.5=0.2.........................equation I

But Z(Second year GPA)= (X-Mean)/S.D.

Here Mean=2.5 and S.D. =0.8.........................equation II

therefore, Z(Second year GPA)= (X-Mean)/S.D. will be

0.2=(X-2.5)/0.8..........................from equation I and II

0.2*0.8=X-2.5

0.16=X-2.5

X=2.66

Hence, 2.66 is second year GPA of someone who has the first year GPA of 2.5

b) Predicting first year GPA of someone who has the second year GPA of 3.0

Standardised value of second year is Z(second year GPA)=(X-Mean)/S.D.

=(3.0-2.5)/0.8

=0.5/0.8

=5/8 S.D.s away from mean

Hence, predicting standardized first year GPA, Z(first year GPA)=correalation* Z(second year GPA)

Z(first year GPA)=0.4*5/8

=2/8

=1/4=0.25.........................equation I

But Z(first year GPA)= (X-Mean)/S.D.

Here Mean=2.1 and S.D. =0.8.........................equation II

therefore, Z(Second year GPA)= (X-Mean)/S.D. will be

0.25=(X-2.1)/0.8..........................from equation I and II

0.25*0.8=X-2.1

0.2=X-2.1

X=2.1+0.2=2.3

X=2.3

Hence, 2.3is first year GPA of someone who has the second year GPA of 3.0

c) Predicting second year GPA of someone who has the first year GPA of 2.1

Standardised value of first year is Z(first year GPA)=(X-Mean)/S.D.

=(2.1-2.1)/0.8

=0/0.8

=0 S.D.s away from mean

i.e. Second year GPA of person will be same as mean of second second year GPA

Hence, 2.5 is second year GPA of someone who has the first year GPA of 2.1

b) Predicting first year GPA of someone who has the unknown second year GPA

first year GPA of someone who has the unknown second year GPA will be mean of first year GPA

Hence, 2.1 is first year GPA of someone who has the unknown second year GPA.

first year GPA second year GPA Mean 2.1 2.5 S.D. 0.8 0.8

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