Scores on an IQ test are normally distributed. A sample of 15 IQ scores had standard deviation s=10. The developer of the test claims that the population standard deviation is =15. Do these data provide sufficient evidence to contradict this claim? Use the =0.05 level of significance.
To determine if the sample data provide sufficient evidence to contradict the developer's claim regarding the population standard deviation, we can use a hypothesis test.
H0: The population standard deviation (σ) = 15
Ha: The population standard deviation (σ) ≠ 15
The first step is to calculate the sample's chi-square statistic (χ²). The chi-square statistic is given by the formula:
χ² = (n - 1) * s² / σ²
Where n is the sample size, s is the sample standard deviation, and σ is the claimed population standard deviation.
In this case, n = 15, s = 10, and σ = 15. Plugging these values into the formula, we get:
χ² = (15 - 1) * 10² / 15²
Simplifying, we have:
χ² = 14 * 100 / 225
χ² = 56 / 9.64
χ² ≈ 5.8
The next step is to find the critical value(s) of the chi-square distribution with degrees of freedom equal to n - 1. Since our significance level (α) is 0.05, we divide α by 2 because it is a two-tailed test (we want to test for both inequality signs). This gives us α/2 = 0.025.
Using a chi-square distribution table or a calculator, we find the critical chi-square value for a significance level of 0.025 with 14 degrees of freedom to be approximately 6.57 (rounded to two decimal places).
Since our calculated chi-square value (5.8) is smaller than the critical chi-square value (6.57), we fail to reject the null hypothesis (H0). This means that there is not enough evidence to contradict the developer's claim that the population standard deviation is 15.
In conclusion, the sample data do not provide sufficient evidence to contradict the claim that the population standard deviation is 15 at the 0.05 level of significance.