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The paired data below consists of test scores and hours of preparation of 5 randomly selected students. The equation of the regression line is y =44.845 + 3.524x and the standard error of estimate is 5.40. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the test.

Hours 5 2 9 6 10
Score 64 48 72 73 80

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The 99% prediction interval for the test score of a person who spent 7 hours preparing for the test is (52.01, 87.02). This means that we are 99% confident that the actual test score will fall within this range.

Calculate the standard error of the prediction:

The standard error of the prediction (SEP) is different from the standard error of estimate (SEE) because it takes into account the uncertainty in the prediction for a single new data point, not just the average error across all data points.

The formula for SEP is:

SEP = SEE * sqrt(1 + (1/n))

where:

SEE is the standard error of estimate (given as 5.40 in this case)

n is the number of data points (5 in this case)

Plugging in the values, we get:

SEP = 5.40 * sqrt(1 + (1/5)) ≈ 6.10

Find the t-statistic:

We need to find the t-statistic for a 99% prediction interval with 5 degrees of freedom (n-1).

You can use a t-table or a calculator to find this value.

For a 99% prediction interval with 5 degrees of freedom, the t-statistic is approximately 2.921.

Calculate the prediction interval:

The prediction interval is calculated using the following formula:

Prediction interval = Predicted score ± t-statistic * SEP

where:

Predicted score is the score predicted by the regression equation for 7 hours of preparation (44.845 + 3.524 * 7 ≈ 69.51)

Plugging in the values, we get:

Prediction interval = 69.51 ± 2.921 * 6.10 ≈ (52.01, 87.02)

Therefore, the 99% prediction interval for the test score of a person who spent 7 hours preparing for the test is (52.01, 87.02).

This means that we are 99% confident that the actual test score will fall within this range.

User Vamsidhar
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