Question

the data show the chest size and weight of several bears. find the regression equation, letting chest size be the independent (x) variable. then find the best predicted weight of a bear with a chest size of 40 inches. is the result close to the actual weight of 382 pounds? use a significance level of 0.05.
chest size (inches) 41 54 44 55 39 51
weight (pounds) 328 528 418 580 296 503
click the icon to view the critical values of the pearson correlation coefficient r.
what is the regression equation?
ŷ = □ + □x (round to one decimal place as needed.)

Explanation:

Step1: Calculate necessary sums

First, we need to calculate the sums of \( x \) (chest size), \( y \) (weight), \( xy \), and \( x^2 \).

Given data:
\( x \): 41, 54, 44, 55, 39, 51
\( y \): 328, 528, 418, 580, 296, 503

\( n = 6 \) (number of data points)

Sum of \( x \): \( \sum x = 41 + 54 + 44 + 55 + 39 + 51 = 284 \)

Sum of \( y \): \( \sum y = 328 + 528 + 418 + 580 + 296 + 503 = 2453 \)

Sum of \( xy \):
\( (41\times328) + (54\times528) + (44\times418) + (55\times580) + (39\times296) + (51\times503) \)
\( = 13448 + 28512 + 18392 + 31900 + 11544 + 25653 = 129449 \)

Sum of \( x^2 \):
\( 41^2 + 54^2 + 44^2 + 55^2 + 39^2 + 51^2 \)
\( = 1681 + 2916 + 1936 + 3025 + 1521 + 2601 = 13679 \)

Step2: Calculate slope (\( b \)) and intercept (\( a \))

The formula for the slope \( b \) of the regression line \( \hat{y} = a + bx \) is:
\( b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \)

The formula for the intercept \( a \) is:
\( a = \frac{\sum y - b\sum x}{n} \)

First, calculate \( b \):
\( b = \frac{6\times129449 - 284\times2453}{6\times13679 - 284^2} \)
\( = \frac{776694 - 696652}{82074 - 80656} \)
\( = \frac{80042}{1418} \approx 56.4 \) (rounded to one decimal place)

Then, calculate \( a \):
\( a = \frac{2453 - 56.4\times284}{6} \)
\( = \frac{2453 - 16017.6}{6} \)
\( = \frac{-13564.6}{6} \approx -2260.8 \) (Wait, that can't be right. Wait, I must have miscalculated. Wait, no, wait: Wait, 56.4*284: 56*284=15904, 0.4*284=113.6, so total 15904+113.6=16017.6. Then 2453 - 16017.6 = -13564.6. Then divide by 6: -13564.6 /6 ≈ -2260.8. But that seems odd. Wait, maybe I made a mistake in the sum of y? Let's recheck the sum of y: 328 + 528 = 856; 856 + 418 = 1274; 1274 + 580 = 1854; 1854 + 296 = 2150; 2150 + 503 = 2653. Oh! I made a mistake in the sum of y. Let's correct that.

Sum of \( y \): 328 + 528 + 418 + 580 + 296 + 503 = 328+528=856; 856+418=1274; 1274+580=1854; 1854+296=2150; 2150+503=2653. So \( \sum y = 2653 \), not 2453. That was the error.

Let's recalculate \( b \) with the correct \( \sum y \):

\( b = \frac{6\times129449 - 284\times2653}{6\times13679 - 284^2} \)
\( = \frac{776694 - 284\times2653}{82074 - 80656} \)
Calculate \( 284\times2653 \): 284*2600=738400, 284*53=15052, so total 738400+15052=753452
So numerator: 776694 - 753452 = 23242
Denominator: 82074 - 80656 = 1418
So \( b = 23242 / 1418 ≈ 16.4 \) (rounded to one decimal place)

Now calculate \( a \):
\( a = \frac{2653 - 16.4\times284}{6} \)
Calculate \( 16.4\times284 \): 16*284=4544, 0.4*284=113.6, so total 4544+113.6=4657.6
Then \( 2653 - 4657.6 = -2004.6 \)
\( a = -2004.6 / 6 ≈ -334.1 \) (Wait, still negative? Wait, maybe I made a mistake in the sum of xy? Let's recheck the sum of xy:

41*328: 41*300=12300, 41*28=1148, total 12300+1148=13448 (correct)
54*528: 50*528=26400, 4*528=2112, total 26400+2112=28512 (correct)
44*418: 40*418=16720, 4*418=1672, total 16720+1672=18392 (correct)
55*580: 55*500=27500, 55*80=4400, total 27500+4400=31900 (correct)
39*296: 40*296=11840, minus 1*296=296, so 11840-296=11544 (correct)
51*503: 50*503=25150, 1*503=503, total 25150+503=25653 (correct)
Sum of xy: 13448+28512=41960; 41960+18392=60352; 60352+31900=92252; 92252+11544=103796; 103796+25653=129449 (correct). So sum of xy is correct.

Wait, maybe the initial sum of y was wrong. Let's add again: 328 + 528 = 856; 856 + 418 = 1274; 1274 + 580 = 1854; 1854 + 296 = 2150; 2150 + 503 = 2653. That's correct.

Wait, maybe the formula for \( a \) is \( \bar{y} - b\bar{x} \), where \( \bar{x} = \sum x / n \), \( \bar{y} = \sum y / n \). Let's try that.

\( \bar{x} = 284 / 6 ≈ 47.333 \)

\( \bar{y} = 2653 / 6 ≈ 442.1667 \)

Then \( a = \bar{y} - b\bar{x} \)

We had \( b ≈ 16.4 \)

So \( a = 442.1667 - 16.4\times47.333 \)

Calculate 16.4*47.333: 16*47.333≈757.328, 0.4*47.333≈18.933, total≈757.328+18.933≈776.261

Then \( a = 442.1667 - 776.261 ≈ -334.094 \), which is approximately -334.1. So that's correct.

Wait, but let's check with a calculator or software. Alternatively, maybe I made a mistake in the slope calculation. Let's recalculate the slope:

\( n\sum xy = 6*129449 = 776694 \)

\( \sum x \sum y = 284*2653 = 753452 \)

Numerator: 776694 - 753452 = 23242

\( n\sum x^2 = 6*13679 = 82074 \)

\( (\sum x)^2 = 284^2 = 80656 \)

Denominator: 82074 - 80656 = 1418

\( b = 23242 / 1418 ≈ 16.4 \) (since 1418*16 = 22688, 23242 - 22688 = 554, 554/1418 ≈ 0.39, so 16.39 ≈ 16.4)

Then \( a = 442.1667 - 16.4*47.333 ≈ 442.1667 - 776.261 ≈ -334.094 ≈ -334.1 \)

So the regression equation is \( \hat{y} = -334.1 + 16.4x \)

Wait, but let's check with one data point. For x=41, \( \hat{y} = -334.1 + 16.4*41 = -334.1 + 672.4 = 338.3 \), which is close to 328 (actual y=328). The difference is 10.3. For x=54, \( \hat{y} = -334.1 + 16.4*54 = -334.1 + 885.6 = 551.5 \), actual y=528, difference 23.5. Hmm, maybe the calculation is correct.

Wait, maybe I made a mistake in the sum of x. Let's add x again: 41 + 54 = 95; 95 + 44 = 139; 139 + 55 = 194; 194 + 39 = 233; 233 + 51 = 284. Correct.

Sum of y: 328 + 528 = 856; 856 + 418 = 1274; 1274 + 580 = 1854; 1854 + 296 = 2150; 2150 + 503 = 2653. Correct.

So the regression equation is \( \hat{y} = -334.1 + 16.4x \)

Answer:

\( \hat{y} = -334.1 + 16.4x \)