Question

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QUESTION 25 4 points Save Answer A guidance counselor at a local high school is interested in determining what, if any, linear relationship there is between high school percentile ranks and college GPAs. A student's percentile rank is calculated by determining the percentage of all students in the graduating class with a final high school GPA at or below his or hers. For example, a student graduating 10th in a class of 300 would have a percentile rank (to one decimal place) of (290/300)x100 = 96.7. The output of the regression fit is shown below. SUMMARY OUTPUT Regression Statistics Multiple R 0.42853296 R Square 0.183640498 Adjusted R Square 0.183443071 Standard Error 0.595165667 Observations 4137 ANOVA Significance F 1.8886E-184 Regression Residual Total of SS MS F 1 329.487 329.487 930. 1704 4135 1464.709 0.354222 4136 1794.196 Coefficients tandard Errt Stat P-value Lower 95% Upper 95% Intercept 1.276896668 0.046049 27.72907 2.3E-155 1.186615795 1.367177542 Percentile Rank 0.017034905 0.000559 30.4987 1.9E-184 0.015939856 0.018129954 The guidance counselor wants to create a 95% prediction interval for the GPA of a student who is graduating with a percentile rank of 80. What is the upper bound of this interval, to three decimal places? Use the fact that the average rank is x=80.8 and with a standard deviation of sy =16.6. Use three (3) decimal places for all calculations and the answer. Be sure to use the prediction interval formula. Hints: Alternative--easier--formulas to the ones presented in the textbook for finding confidence intervals and prediction intervals Confidence Interval Ita/2. 1. ev n 1- +- (x*-x) (n-1) Prediction Interval Vita/2.n-25e, 1+ 1 n + (x+ - x)? (n-1) where se is the "regression standard error and Xp=x* is the particular value of x that you are using to make the interval for Y. Use three decimal places for all calculations up to the answer. Use the prediction interval formula, not the confidence interval formula.

Answer 1

as we are instructed to use all calculations rounded off to 3 decimal places, i am using all the data values in the output by rounding off to decimal places.

the regression equation is:-

y = 1.277 + 0.17x

the predicted value of y for percentile rank (x) = 80 is:-

$\hat{y}$ = 1.277+(0.17*80) = 14.877

t critical value for df= (n-2 )= (4137-2) = 4135 ,alpha=0.05,both tailed test be:-

$t_{\frac{\alpha}{2},n-2}=1.961$

the necessary data are:-

standard error of regression ($s_{e}$) = 0.595

$\bar{x}=80.8,x^*=80,s_{x}=16.6,n=4137,\hat{y}=14.877$

the 95% prediction interval is :-

1 9+tan-2 + be + 11+-+ n (x* -T2 (n-1)s

= 14.877 +1.961 * 0.595* 1 (80-80.812 1+, 4137"(4137-1)* 16.62

= (13.710.16.044)

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