Regression and Predictions
Exercises 13–28 use the same ...

Question

Regression and Predictions

Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1.

Find the regression equation, letting the first variable be the predictor (x) variable.

Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5.

Subway and the CPI Use the subway/CPI data from the preceding exercise. What is the best predicted value of the CPI when the subway fare is $3.00?

Accepted Answer

Hello, everyone. Let's take a look at this question together. The table below shows the number of hours students studied for a final exam and their corresponding exam scores, which are out of 100. Find the best predicted exam score for a student who studied for 9 hours. And here we have our table of the number of hours students studied for a final exam and their corresponding exam scores, and we have to determine is the best predicted exam score for a student who studied for 9 hours, answer choice A 90, answer choice B 87, answer choice C 95, or answer choice D 83. So the first step in solving this problem is to identify the predictor and Response variables, which we know the predictor variable or X is the study hours, and the response variable or Y is the exam score. So then the next step is to compute the least squares regression equation, which is Y is equal to B0 plus B1X. And using the given data, we can compute the following values, which gives us The mean of X or the mean of the study hours which we can see is equal to 6, and the mean of Y or the mean of the exam scores is equal to 73.5. So the value of 6 represents X bar and the value of 73.5 represents Y bar, and then we can calculate the standard deviation of X and Y using the formulas for the standard. of X and the standard deviation of Y and plugging in the values that we computed, for the standard deviation of X, we calculate 2.90 hours, and for the standard deviation of Y, we calculate 13.34 points. And we can also calculate the correlation coefficient R using the formula for the correlation coefficient, which we calculate as 0.99. 4. And using the values that we calculated for the standard deviation of X and Y, as well as the correlation coefficient and the values for X and Y, we can calculate our least squares regression equation of Y equals B0 plus B1X by calculating the values for B0 and B1, and the formula for B1 is R multiplied by the. Standard deviation of Y divided by the standard deviation of X, and plugging in the values we calculate 4.57 as our value for B1, and then for B0, we use the formula for B0, which is Y minus B1 multiplied by X bar, and plugging in the values we calculate 46.1 as the value for B0, as well as we know the value for X. is equal to 9, and as a result, the regression equation is equal to Y is equal to 46.1 plus 4.57 multiplied by X, and using the regression equation, we can plug in the value for X, which gives us the formula of Y hat is equal to 46.1 plus in parentheses 4.57 multiplied by 9, which is equal to 87. Therefore, the predicted exam score is 87, and looking at the original table, knowing we have a predicted exam score of 87 for a student who studied for 9 hours, since our study hours do not show a student who studied for exactly 9 hours, we can use. The value for the student who studied for 10 hours, who had an exam score of 90, so therefore, the closest actual score is 90, which makes answer choice A the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

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Key Concepts

Regression Analysis

Predictor and Response Variables

Prediction Procedure

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