To answer this question, we can perform a one-tailed chi-square test.
Step 1: To start, we can set up the hypothesis as follows:
- Null hypothesis (H0): The population standard deviation is equal to 0.12
- Alternative hypothesis (H1): The population standard deviation is less than 0.12
Step 2: We are given the following parameters:
- Sample standard deviation (s) = 0.112
- Sample size (n) = 28
- Population standard deviation (σ) = 0.12
- Level of significance (α) = 0.05
Step 3: Now, we calculate the chi-square statistic. We can use the formula (n-1)*s²/σ².
Substituting the given values into the formula, we get:
(28 - 1) * (0.112)^2 / (0.12)^2 = 23.52
Step 4: Next, we need to determine the critical value from chi-square distribution table. Here, our degrees of freedom would be n - 1 = 28 - 1 = 27 and we're conducting a one-tailed test at a 5% level of significance. Therefore, referring to the chi-square distribution table for 27 degrees of freedom, we get a critical value as 40.113
Step 5: Finally, for the decision rule, we compare our chi-square statistic with the critical value. The decision rule for a left-tailed test is: If the calculated chi-square statistic is less than the critical value from the chi-square distribution, we reject the null hypothesis.
Here, our calculated chi-square statistic 23.52 is less than the critical value 40.113. Hence, we reject the null hypothesis.
Therefore, we can conclude at the 5% level of significance that the population standard deviation is statistically less than 0.12.