A sample of 30 trees has a standard deviation of 0.8m. Test the hypothesis that this result is less than the theoretical value of 0.6m, using a 5% significance level. Calculate the test statistic and outcome of the test.
To test the hypothesis, we will use a one-sample t-test.
The null hypothesis (H0) states that the population standard deviation is equal to 0.6m.
The alternative hypothesis (H1) states that the population standard deviation is less than 0.6m.
Given that the sample size (n) is 30, the sample standard deviation (s) is 0.8m, and the significance level (α) is 5%, we can calculate the test statistic using the formula:
t = (s - σ) / (s/√n),
where σ is the assumed population standard deviation (0.6m) and √n is the square root of the sample size (√30 ≈ 5.477).
Substituting the values:
t = (0.8 - 0.6) / (0.8 / 5.477)
t = 0.2 / (0.8 / 5.477)
t ≈ 0.2195
To determine the outcome of the test, we compare the test statistic with the critical value from the t-distribution table at α = 5% significance level (one-tailed test).
Since the alternative hypothesis states that the population standard deviation is less than 0.6m, we need to find the critical value from the t-distribution that corresponds to a probability of 5% in the lower tail.
Based on the degrees of freedom (n-1 = 30-1 = 29) and a significance level of 5%, the critical value is approximately -1.6991.
Since the test statistic (0.2195) is greater than the critical value (-1.6991), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to conclude that the sample standard deviation is less than the theoretical value of 0.6m at a 5% significance level.
To test the hypothesis, we can use the one-sample t-test. The test statistic can be calculated using the formula:
t = (x̄ - μ) / (s / √n)
Where:
x̄ is the sample mean
μ is the population mean (theoretical value)
s is the sample standard deviation
n is the sample size
Given:
Sample size (n) = 30
Sample standard deviation (s) = 0.8m
Population mean (μ) = 0.6m
Significance level α = 0.05
Step 1: Calculate the test statistic (t-value)
t = (x̄ - μ) / (s / √n)
Step 2: Determine the critical value
Since we are testing if the result is less than the theoretical value, this is a one-tailed test. The critical value for a one-tailed test with a significance level of 0.05 is obtained from the t-distribution table or calculator.
For a sample size of 30 and a significance level of 0.05, the critical value is approximately -1.699.
Step 3: Compare the test statistic with the critical value
If the test statistic is less than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Let's calculate the test statistic and compare it with the critical value:
t = (x̄ - μ) / (s / √n)
t = (0.8 - 0.6) / (0.8 / √30)
t = 0.2 / (0.8 / √30)
t ≈ 0.2 / 0.146
t ≈ 1.37
Since the calculated t-value of 1.37 is greater than the critical value of -1.699, we fail to reject the null hypothesis.
The outcome of the test is that there is not enough evidence to support the claim that the result (standard deviation of 0.8m) is less than the theoretical value of 0.6m at a 5% significance level.