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Find the regression​ equation, letting the diameter be the predictor​ (x) variable. Find the best predicted circumference of a

beachballbeachball
with a diameter of 42.5
cm. How does the result compare to the actual circumference of 133.5
​cm? Use a significance level of 0.05.
Baseball
D: 7.3
C: 22.9
Basketball
D: 23.6
C: 74.1
Golf
D: 4.3
C: 13.5
Soccer
D: 22.1
C: 69.4
Tennis
D: 7.1
C: 22.3
Ping-Pong
D: 3.9
C: 12.3
Volleyball
D: 21.2
C: 66.6
The regression equation is
y= _____ + _____x
The best predicted circumference for a diameter of 42.5 cm is _____ cm.
How does the result compare to the actual circumference of 133.5 cm?
a) Since 42.5 cm is within the scope of the sample​ diameters, the predicted value yields the actual circumference.
b) Even though 42.5 cm is beyond the scope of the sample​ diameters, the predicted value yields the actual circumference.
c) Since 42.5 cm is beyond the scope of the sample​ diameters, the predicted value yields a very different circumference.
d) Even though 42.5 cm is within the scope of the sample​ diameters, the predicted value yields a very different circumference.

User Steve Dunn
by
7.9k points

1 Answer

1 vote

Answer:

Explanation:

Using the given data, we can calculate the regression equation as follows:

First, calculate the means of x and y:

mean(x) = (7.3 + 23.6 + 4.3 + 22.1 + 7.1 + 3.9 + 21.2) / 7 = 13.1143

mean(y) = (22.9 + 74.1 + 13.5 + 69.4 + 22.3 + 12.3 + 66.6) / 7 = 42.3286

Then, calculate the sample standard deviations of x and y:

s_x = sqrt(((7.3 - 13.1143)^2 + (23.6 - 13.1143)^2 + ... + (21.2 - 13.1143)^2) / 6) = 7.8836

s_y = sqrt(((22.9 - 42.3286)^2 + (74.1 - 42.3286)^2 + ... + (66.6 - 42.3286)^2) / 6) = 25.3612

Calculate the sample covariance:

cov(x,y) = ((7.3 - 13.1143)(22.9 - 42.3286) + (23.6 - 13.1143)(74.1 - 42.3286) + ... + (21.2 - 13.1143)*(66.6 - 42.3286)) / 6 = 194.4714

Calculate the slope of the regression line:

b = cov(x,y) / s_x^2 = 194.4714 / (7.8836^2) = 3.1343

Calculate the intercept of the regression line:

a = mean(y) - b * mean(x) = 42.3286 - 3.1343 * 13.1143 = 1.1639

Therefore, the regression equation is:

y = 1.1639 + 3.1343x

To find the best predicted circumference for a diameter of 42.5 cm, we substitute x = 42.5 into the regression equation:

y = 1.1639 + 3.1343(42.5) = 133.262

The predicted circumference is 133.262 cm.

To determine whether the predicted value is significantly different from the actual circumference of 133.5 cm, we can calculate the standard error of the estimate:

s_e = sqrt(((22.9 - 133.262)^2 + (74.1 - 133.262)^2 + ... + (66.6 - 133.262)^2) / 5) = 17.2538

Using a significance level of 0.05 and a t-distribution with 5 degrees of freedom (n - 2), the critical t-value is 2.571. The margin of error is then:

ME = t_critic * s_e = 2.571 * 17.2538 = 44.3726

Since the actual circumference of 133.5 cm falls within the margin of error (133.262 - 44.3726 to 133.262 + 44.3726), we can conclude that the predicted value is not significantly different from the actual circumference. Therefore, the answer is:

b) Even though 42.5 cm is beyond the scope of the sample​ diameters, the predicted value yields the actual circumference.

User Fiiv
by
7.7k points