To test whether the engineer is correct, we will conduct a hypothesis test. The null hypothesis (H0) is that the average weight of trees is 706 pounds, while the alternative hypothesis (Ha) is that the average weight of trees is different from 706 pounds.
We will use a significance level (α) of 0.05.
First, let's calculate the sample mean and sample standard deviation:
Sample mean (x̄) = (5,109 + 8,409 + 16,209) / 3 = 9,242.33 pounds
Sample standard deviation (s) = sqrt(((5,109 - 9,242.33)^2 + (8,409 - 9,242.33)^2 + (16,209 - 9,242.33)^2) / 2) = 5,151.22 pounds
Next, we will calculate the test statistic, which follows a t-distribution for small samples:
t = (x̄ - μ) / (s / sqrt(n))
n = 3 (number of observations), μ = 706 (null hypothesis mean), s = 5,151.22 (sample standard deviation)
t = (9,242.33 - 706) / (5,151.22 / sqrt(3))
t = 8,536.33 / 2,974.27
t = 2.868
Using a t-table or calculator, we find that the critical t-value for a two-tailed test at α = 0.05 with 2 degrees of freedom (n-1) is approximately ± 4.303.
Since the calculated t-value (2.868) does not exceed the critical t-value (± 4.303), we fail to reject the null hypothesis.
Therefore, based on the sample data, we do not have sufficient evidence to conclude that the average weight of the trees is different from 706 pounds.
An engineer believes the average weight of a tree is 706 pounds. He feels it's safe to assume the weights are normally distributed. He weighs three randomly selected trees and records their weight: 5,109 pounds 8,409 pounds 16,209 pounds Test whether the engineer is correct
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