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.

Answer 1

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.