Question

1. the total number, y, of people in a grocery store x hours after it opens is shown in this table.

hours	total people
2	9
3	15
4	25
5	22
6	30

the data is modeled with a best - fit linear function. which total number of people was closest to the predicted value for that hour?

a. 5
b. 9
c. 22
d. 30

Explanation:

Step1: Calculate Mean of x and y

First, find the mean of \( x \) (hours) and \( y \) (total people).
\( x \)-values: \( 1, 2, 3, 4, 5, 6 \)
Mean of \( x \): \( \bar{x} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5 \)
\( y \)-values: \( 5, 9, 15, 25, 22, 30 \)
Mean of \( y \): \( \bar{y} = \frac{5 + 9 + 15 + 25 + 22 + 30}{6} = \frac{106}{6} \approx 17.67 \)

Step2: Calculate Slope (m) of Best-Fit Line

Slope formula: \( m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)
Calculate \( (x_i - \bar{x})(y_i - \bar{y}) \) for each \( i \):
- \( (1 - 3.5)(5 - 17.67) = (-2.5)(-12.67) = 31.675 \)
- \( (2 - 3.5)(9 - 17.67) = (-1.5)(-8.67) = 13.005 \)
- \( (3 - 3.5)(15 - 17.67) = (-0.5)(-2.67) = 1.335 \)
- \( (4 - 3.5)(25 - 17.67) = (0.5)(7.33) = 3.665 \)
- \( (5 - 3.5)(22 - 17.67) = (1.5)(4.33) = 5.995 \)
- \( (6 - 3.5)(30 - 17.67) = (2.5)(12.33) = 30.825 \)
Sum of these: \( 31.675 + 13.005 + 1.335 + 3.665 + 5.995 + 30.825 = 86.5 \)

Calculate \( (x_i - \bar{x})^2 \) for each \( i \):
- \( (1 - 3.5)^2 = 6.25 \)
- \( (2 - 3.5)^2 = 2.25 \)
- \( (3 - 3.5)^2 = 0.25 \)
- \( (4 - 3.5)^2 = 0.25 \)
- \( (5 - 3.5)^2 = 2.25 \)
- \( (6 - 3.5)^2 = 6.25 \)
Sum of these: \( 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5 \)

Slope \( m = \frac{86.5}{17.5} \approx 4.943 \)

Step3: Calculate Y-Intercept (b)

Using \( \bar{y} = m\bar{x} + b \):
\( 17.67 = 4.943(3.5) + b \)
\( 17.67 = 17.3005 + b \)
\( b \approx 17.67 - 17.3005 = 0.3695 \)

Best-fit line: \( y = 4.943x + 0.3695 \)

Step4: Predict y for Each x and Find Residuals

- For \( x = 1 \): \( y = 4.943(1) + 0.3695 \approx 5.31 \); Residual: \( |5 - 5.31| = 0.31 \)
- For \( x = 2 \): \( y = 4.943(2) + 0.3695 \approx 10.25 \); Residual: \( |9 - 10.25| = 1.25 \)
- For \( x = 5 \): \( y = 4.943(5) + 0.3695 \approx 25.08 \); Residual: \( |22 - 25.08| = 3.08 \)
- For \( x = 6 \): \( y = 4.943(6) + 0.3695 \approx 30.03 \); Residual: \( |30 - 30.03| = 0.03 \) Wait, but wait—wait, the options are A (5), B (9), C (22), D (30). Wait, maybe miscalculation? Wait, no—wait, the best-fit line is linear, but let's check the residual for each option:

Wait, actually, the residual is the absolute difference between actual \( y \) and predicted \( y \). Let's recalculate predictions:

For \( x=1 \): \( y = 4.943(1) + 0.37 \approx 5.31 \), residual \( |5 - 5.31| = 0.31 \)
For \( x=2 \): \( y = 4.943(2) + 0.37 \approx 10.26 \), residual \( |9 - 10.26| = 1.26 \)
For \( x=5 \): \( y = 4.943(5) + 0.37 \approx 25.09 \), residual \( |22 - 25.09| = 3.09 \)
For \( x=6 \): \( y = 4.943(6) + 0.37 \approx 30.03 \), residual \( |30 - 30.03| = 0.03 \). But wait, the option D is 30. But wait, maybe I made a mistake in slope? Wait, alternatively, maybe the best-fit line is calculated with two points (e.g., first and last). Let's try that: points (1,5) and (6,30). Slope \( m = \frac{30 - 5}{6 - 1} = 5 \). Then equation: \( y - 5 = 5(x - 1) \) → \( y = 5x \). Then predictions:

- \( x=1 \): \( y=5 \), residual 0
- \( x=2 \): \( y=10 \), residual 1
- \( x=3 \): \( y=15 \), residual 0 (wait, actual is 15, so residual 0)
- \( x=4 \): \( y=20 \), residual 5 (actual 25)
- \( x=5 \): \( y=25 \), residual 3 (actual 22)
- \( x=6 \): \( y=30 \), residual 0

But the problem says "best-fit linear function" (not necessarily through first and last). But the residual for \( x=6 \) (D:30) is very small (0.03 with the first calculation, 0 with the two-point line). But wait, the options are A (5), B (9), C (22), D (30). Wait, but when \( x=6 \), actual is 30, predicted is ~30.03, so residual ~0.03. For \( x=1 \), predicted ~5.31, residual ~0.31. For \( x=2 \), predicted ~10.25, residual ~1.25. For \( x=5 \), predicted ~25.08, residual ~3.08. So the smallest residual is for D (30) or A (5). Wait, but maybe the best-fit line is calculated with all points. Let's recalculate the slope correctly:

Sum of \( x_i y_i \): \( (1)(5) + (2)(9) + (3)(15) + (4)(25) + (5)(22) + (6)(30) = 5 + 18 + 45 + 100 + 110 + 180 = 458 \)
Sum of \( x_i^2 \): \( 1 + 4 + 9 + 16 + 25 + 36 = 91 \)
Slope \( m = \frac{n\sum x_i y_i - \sum x_i \sum y_i}{n\sum x_i^2 - (\sum x_i)^2} \)
\( n=6 \), \( \sum x_i=21 \), \( \sum y_i=106 \)
\( m = \frac{6(458) - 21(106)}{6(91) - 21^2} = \frac{2748 - 2226}{546 - 441} = \frac{522}{105} \approx 4.971 \)
Y-intercept \( b = \frac{\sum y_i - m\sum x_i}{n} = \frac{106 - 4.971(21)}{6} = \frac{106 - 104.391}{6} \approx \frac{1.609}{6} \approx 0.268 \)
So equation: \( y = 4.971x + 0.268 \)

Predictions:
- \( x=1 \): \( 4.971 + 0.268 \approx 5.24 \), residual \( |5 - 5.24| = 0.24 \)
- \( x=2 \): \( 9.942 + 0.268 \approx 10.21 \), residual \( |9 - 10.21| = 1.21 \)
- \( x=5 \): \( 24.855 + 0.268 \approx 25.12 \), residual \( |22 - 25.12| = 3.12 \)
- \( x=6 \): \( 29.826 + 0.268 \approx 30.09 \), residual \( |30 - 30.09| = 0.09 \)

So the smallest residual is for D (30), with residual ~0.09, followed by A (5) with ~0.24. But wait, the problem says "which total number of people was closest to the predicted value". So D (30) has the smallest residual. Wait, but maybe I made a mistake. Wait, let's check the actual data: at x=6, actual is 30, predicted is ~30.09, so very close. At x=1, actual 5, predicted ~5.24, close but not as close as 30. So the answer is D.

Answer:

D. 30