To solve this problem, we need to find the least-squares regression line for the given data and use it to predict the exam score for a student who slept for 7 hours.

Data:

• Hours slept (x): 3, 5, 2, 8, 2, 4, 4, 5, 6, 3
• Exam scores (y): 65, 80, 60, 88, 66, 78, 85, 90, 90, 71

Steps:

1. Calculate the least-squares regression line using the formula: $\hat{y} = b_0 + b_1x$ where:

   • $b_1 = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$
   • $b_0 = \frac{\sum y - b_1 \sum x}{n}$

2. Find the predicted score for a student who slept 7 hours by substituting $x = 7$ into the regression equation.

3. Interpret the reasonableness of this prediction based on the given data.

Let’s perform the calculations to get the regression equation.

Calculations:

1. Sum of x values:
   $\sum x = 3 + 5 + 2 + 8 + 2 + 4 + 4 + 5 + 6 + 3 = 42$

2. Sum of y values:
   $\sum y = 65 + 80 + 60 + 88 + 66 + 78 + 85 + 90 + 90 + 71 = 707$

3. Sum of xy values:
   $\sum xy = (3 \times 65) + (5 \times 80) + (2 \times 60) + (8 \times 88) + (2 \times 66) + (4 \times 78) + (4 \times 85) + (5 \times 90) + (6 \times 90) + (3 \times 71) = 2732$

4. Sum of x² values:
   $\sum x^2 = (3^2) + (5^2) + (2^2) + (8^2) + (2^2) + (4^2) + (4^2) + (5^2) + (6^2) + (3^2) = 134$

5. Number of data points (n):
   $n = 10$

Now, applying the formulas for $b_1$ and $b_0$:

$b_1 = \frac{10(2732) - (42)(707)}{10(134) - (42)^2} = \frac{27320 - 29694}{1340 - 1764} = \frac{-2380}{-424} = 5.04$

$b_0 = \frac{707 - (5.04)(42)}{10} = \frac{707 - 211.68}{10} = \frac{495.32}{10} = 56.11$

Thus, the least-squares regression line is: $\hat{y} = 5.04x + 56.11$

Predicted score for 7 hours of sleep:

Substitute $x = 7$ into the regression equation:

$\hat{y} = 5.04(7) + 56.11 = 35.28 + 56.11 = 91.39$

Rounding to the nearest whole number, the predicted score is 91.

Is this prediction reasonable?

The model is based on data from 2 to 8 hours of sleep. A prediction for 7 hours is within this range, so it is reasonable.

Final Answer:

The correct answer is:

B. ( \hat{y} = 5.04x + 56.11; 91; Yes, it is reasonable.

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Would you like more details or further clarification on any step? Here are some related questions to explore:

1. How do we interpret the slope and y-intercept of the regression line?
2. What is the formula for calculating the correlation coefficient, and how can it be used to assess the fit of the regression model?
3. What are the limitations of using regression models to predict outcomes outside of the observed data range (extrapolation)?
4. How can we determine the residuals for this regression model, and what do they tell us about the accuracy of our predictions?
5. How can we perform a hypothesis test to check the significance of the regression slope?

Tip: Always check whether the data you're predicting for falls within the observed range, as extrapolating too far can lead to unreliable predictions.