Predicting GPA from Standardized Test Scores Using Regression Analysis

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Topic 7 Homework (Nonadaptive) Question 9 of 16 (1 point)|Question Attempt: 3 of Unlimited

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Question 9 Suppose we are examining the relationship between scores on a popular standardized test and performance in college. We have chosen a random sample of 45 students just finishing their first year of college, and for each student we've recorded her score on this standardized test (from 400 to 1600) and her grade point average (on a four-point scale) for her first year in college. Letting x denote score on the standardized test and y denote first-year college grade point average, the least-squares regression equation computed from the data is =y+0.90320.0015x. We're now interested in predicting the first-year grade point average of a student who scored 1010 on the standardized test. We're also interested in a prediction interval for this grade point average and a confidence interval for the mean first-year grade point average of students who scored 1010 on the standardized test. We have already computed the following for our data.

mean square error ≈MSE 0.326 ≈+145−1010x2Σ=i145−xix2 0.0376, where x1, x2, ..., x45 denote the standardized test scores in the sample, and x denotes their mean Based on this information, and assuming that the regression assumptions hold, answer the questions below.

(If necessary, consult a list of formulas.)

(a)What is the 99% prediction interval for an individual value for grade point average when the standardized test score is 1010? (Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places.)

Lower limit:

Upper limit:

(b)Consider (but do not actually compute) the 99% confidence interval for the mean grade point average when the standardized test score is 1010. How would this confidence interval compare to the prediction interval computed above (assuming that both intervals are computed from the same sample data)?

The confidence interval would be identical to the prediction interval. The confidence interval would be positioned to the right of the prediction interval. The confidence interval would have the same center as, but would be wider than, the prediction interval. The confidence interval would be positioned to the left of the prediction interval. The confidence interval would have the same center as, but would be narrower than, the prediction interval.

(c)For the standardized test score values in this sample, 1310 is more extreme than 1010 is, that is, 1310 is farther from the sample mean standardized test score than 1010 is. How would the 99% prediction interval for the mean grade point average when the standardized test score is 1010 compare to the 99% prediction interval for the mean grade point average when the standardized test score is 1310?

The interval computed from a standardized test score of 1010 would be narrower but have the same center.

The interval computed from a standardized test score of 1010 would be wider but have the same center.

The intervals would be identical.

The interval computed from a standardized test score of 1010 would be narrower and have a different center.

The interval computed from a standardized test score of 1010 would be wider and have a different center.

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Solution

Let's go through each part of this question carefully.

Part (a)

We are given:

The least-squares regression equation: $\hat{y} = 0.9032 + 0.0015x$
$x = 1010$ (the score we’re using for prediction)
$\text{MSE} = 0.326$
$\sum_{i=1}^{45} (x_i - \bar{x})^2 = 0.0376$

To calculate the 99% prediction interval for an individual GPA when the standardized test score is 1010, we will:

Predict the GPA using $x = 1010$ .
Calculate the standard error for a prediction interval.
Use the t-distribution with a 99% confidence level.

Step 1: Prediction using $x = 1010$

Substitute $x = 1010$ into the regression equation: $\hat{y} = 0.9032 + (0.0015)(1010) = 0.9032 + 1.515 = 2.4182$

Step 2: Standard Error for Prediction

The standard error $SE$ for a prediction interval is given by: $SE = \sqrt{\text{MSE} \left(1 + \frac{1}{n} + \frac{(x - \bar{x})^2}{\sum_{i=1}^{n} (x_i - \bar{x})^2}\right)}$ Using the provided values:

$\text{MSE} = 0.326$
$n = 45$
$(x - \bar{x})^2 / \sum_{i=1}^{45} (x_i - \bar{x})^2 = 0.0376$

Plugging these in: $SE = \sqrt{0.326 \left(1 + \frac{1}{45} + 0.0376\right)}$

Step 3: Find $t$ -value

For a 99% confidence level with $n - 2 = 43$ degrees of freedom, we find the critical $t$ -value, which is approximately 2.695.

Step 4: Prediction Interval

The prediction interval is: $\hat{y} \pm t \cdot SE$ After calculating $SE$ , substitute in $\hat{y} = 2.4182$ and $t = 2.695$ to find the interval.

Part (b)

The confidence interval for the mean GPA (not an individual prediction) when the test score is 1010 would be narrower than the prediction interval. This is because a confidence interval for the mean has a smaller error term than a prediction interval, which must account for individual variation.

Part (c)

For a standardized test score of 1310, which is farther from the mean than 1010:

The prediction interval for 1010 would likely be narrower because it is closer to the mean.

The correct answer is:

"The interval computed from a standardized test score of 1010 would be narrower and have a different center."

Would you like the calculations for the standard error or any other steps?

Relative Questions

How would the prediction interval change if the MSE were lower?
Why does the prediction interval include more variability than the confidence interval?
How does the sample size affect the width of the confidence interval?
What assumptions underlie the use of a linear regression model in this context?
How does the selection of a 99% confidence level affect the intervals compared to a 95% level?

Tip:

Always distinguish between prediction intervals (for individual predictions) and confidence intervals (for the mean prediction) when interpreting regression outputs.

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Math Problem Analysis

Mathematical Concepts

Statistics
Regression Analysis
Prediction and Confidence Intervals

Formulas

Least-squares regression equation y = b0 + b1 * x
Prediction interval: y ± t * SE
Standard error for prediction interval SE = sqrt(MSE * (1 + 1/n + (x - x̄)^2 / Σ(xi - x̄)^2))

Theorems

Least-Squares Regression
Confidence Interval
Prediction Interval

Suitable Grade Level

Undergraduate Statistics

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