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
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.
To determine whether the engineer's belief about the average weight of a tree is supported by the data, we can perform a hypothesis test. Let's go through the steps:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha):
- Null hypothesis (H0): The average weight of the tree is 706 pounds.
- Alternative hypothesis (Ha): The average weight of the tree is not 706 pounds.
Step 2: Determine the significance level (α):
- This is a subjective choice, but for this example, let's assume a significance level of α = 0.05. This means we want the probability of making a Type I error (rejecting the null hypothesis when it is true) to be 5%.
Step 3: Collect and summarize the data:
- The engineer weighed three randomly selected trees and recorded their weights: 5,109 pounds, 8,409 pounds, and 16,209 pounds.
Step 4: Calculate the test statistic:
- Since the sample size is small (n = 3) and we don't know the population standard deviation, we'll use a t-test.
- The test statistic is calculated using the formula: t = (x̄ - μ) / (s / √n), where x̄ is the sample mean, μ is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
- Calculating the sample mean: x̄ = (5,109 + 8,409 + 16,209) / 3 = 9,242.33 pounds.
- Calculating the sample standard deviation: s = √[((5,109 - 9,242.33)² + (8,409 - 9,242.33)² + (16,209 - 9,242.33)²) / (3 - 1)] ≈ 4,998.08 pounds.
- Calculating the test statistic: t = (9,242.33 - 706) / (4,998.08 / √3) ≈ 6.06.
Step 5: Determine the critical value or p-value:
- Since this is a two-tailed test, we need to find the critical t-value(s) or p-value.
- The degrees of freedom (df) is equal to n - 1 = 2.
- Looking up the critical t-value using a t-distribution table or calculator with α = 0.05 and df = 2 results in t_critical = ±4.30 (approximately).
- Alternatively, we can calculate the p-value associated with the test statistic. Using a t-distribution calculator, the p-value is approximately 0.046 (two-tailed).
Step 6: Make a decision and interpret the results:
- If the absolute value of the test statistic is greater than the critical t-value or the p-value is less than α, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
- In this case, |6.06| > 4.30 and the p-value (0.046) < α (0.05), so we reject the null hypothesis.
- This means that based on the given data, there is sufficient evidence to conclude that the average weight of trees is not 706 pounds.
In summary, the engineer's belief is not supported by the data, and we reject the hypothesis that the average weight of trees is 706 pounds.