In Professor White’s statistics course the correlation between the students’ total scores before the final examination and their final examination scores is r = 0.9. The pre-exam totals for all students in the course have mean 275 and standard deviation 50. The final exam scores have mean 70 and standard deviation 10. Professor White has lost Mary’s total scores before the exam but knows her final exam was 81. He decides to predict her pre-exam total from her final score.
4-1) (2 marks) What is the slope of the least-squares regression line of pre-exam total scores on final exam scores in this course? What is the intercept?
4-2) (1 mark) Use the regression line to predict Mary’s pre-exam total scores.
4-3) (2 marks) Mary doesn’t think this method accurately predicts how well she did on the pre-exam. Calculate r2 and use the value you get to argue that her actual score could have been much higher or much lower than the predicted value.
Suppose there are two defective items in a pool of five items. You want to test one
item at a time until you identify the second defective item. Find following probabilities
a. You need four tests. (i.e. The second defective item will be found at the fourth
test.)
b. You need at most three tests.
c. Given the second test found at the third test, what is the probability that you
found the first defective at the first test?
To find the slope and intercept of the least-squares regression line, we can use the formula:
slope (b) = r * (SDy / SDx)
intercept (a) = mean of y - (b * mean of x)
where r is the correlation coefficient, SDy is the standard deviation of the y variable (final exam scores), SDx is the standard deviation of the x variable (pre-exam total scores), and the mean of x and y are the means of the respective variables.
Given:
r = 0.9
SDx = 50
mean of x = 275
SDy = 10
mean of y = 70
4-1) To find the slope:
slope (b) = 0.9 * (10/50) = 0.9 * 0.2 = 0.18
To find the intercept:
intercept (a) = 70 - (0.18 * 275) = 70 - 49.5 = 20.5
4-2) To predict Mary's pre-exam total scores using the regression line:
Mary's final exam score (y) = 81
predicted pre-exam total score (x) = a + b * y
= 20.5 + 0.18 * 81
= 20.5 + 14.58
= 35.08
Therefore, the predicted pre-exam total score for Mary is approximately 35.08.
4-3) To calculate r^2, which represents the proportion of the total variation in y that can be explained by the regression model:
r^2 = r^2
Given that r = 0.9, r^2 = 0.9^2 = 0.81
This means that 81% of the variation in the final exam scores can be explained by the pre-exam total scores.
Since r^2 = 0.81, 81% of the variation in Mary's final exam score (81) can be explained by her pre-exam total score using the regression model. However, this also implies that 19% of the variation is unexplained.
Thus, her actual pre-exam score could be either higher or lower than the predicted value (35.08) because there is still some variation that is not captured by the regression model.