A state school administrator says that the standard deviation of test scores for 8th grade students who took a U.S. history assessment test is less than 30 points. You work for the administrator and are asked to test this claim. You randomly select 18 tests and find that the tests have a standard deviation of 33.6 points. At a = 0.01, is there enough evidence to support the administrator's claim?
To test the administrator's claim, we can use a one-sample t-test. The null hypothesis (H0) would be that the standard deviation of the test scores is equal to 30 points, while the alternative hypothesis (Ha) would be that the standard deviation is less than 30 points.
Calculating the test statistic:
Sample size (n) = 18
Sample standard deviation (s) = 33.6
Hypothesized standard deviation (σ) = 30
Degrees of freedom (df) = n - 1 = 18 - 1 = 17
The test statistic (t) is calculated as:
t = (s - σ) / (s / √n)
t = (33.6 - 30) / (33.6 / √18)
t ≈ 1.299
The critical value for a one-tailed t-test at a significance level of α = 0.01 and 17 degrees of freedom is -2.898 (obtained from t-table or calculator).
Since the test statistic (1.299) is greater than the critical value (-2.898), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to support the administrator's claim that the standard deviation of test scores for 8th-grade students who took the U.S. history assessment test is less than 30 points.
To test the administrator's claim, we can use a hypothesis test. We'll set up the null and alternative hypotheses as follows:
Null Hypothesis (H0): The standard deviation of test scores is equal to or greater than 30 points (σ >= 30).
Alternative Hypothesis (Ha): The standard deviation of test scores is less than 30 points (σ < 30).
Next, we'll calculate the test statistic using the chi-square distribution. The test statistic for a hypothesis test for the population standard deviation is:
χ² = (n - 1) * s² / σ²
where n is the sample size, s is the sample standard deviation, and σ is the assumed population standard deviation.
In this case, n = 18, s = 33.6, and σ = 30. Plugging in these values, we have:
χ² = (18 - 1) * (33.6)² / (30)²
Calculating this expression, we find:
χ² ≈ 27.116
To determine the critical χ² value at a = 0.01 with (n - 1) degrees of freedom, we'll consult a chi-square distribution table or use statistical software. For (n - 1) = 17 degrees of freedom and α = 0.01, the critical χ² value is approximately 31.410.
Since the calculated χ² value (27.116) is less than the critical χ² value (31.410), we fail to reject the null hypothesis. Therefore, there is not enough evidence to support the administrator's claim that the standard deviation of test scores is less than 30 points.