The IQ’s of 51 students are tested and it is found that the average IQ is 108 with a standard deviation of 10. Test H0:µ^2 = 81 vs Ha:µ^2 not =81 using a 5% significance level. Calculate the test statistic and outcome of the test
To test the hypothesis H0:µ^2 = 81 vs Ha:µ^2 ≠ 81, we can use the test statistic for a sample variance:
t = [(n-1)s^2] / σ^2
where n is the sample size, s^2 is the sample variance, and σ^2 is the hypothesized population variance.
Given that we have a sample size of 51, an average IQ of 108, and a standard deviation of 10, we can calculate the sample variance:
s^2 = (10^2) / (51-1)
s^2 = 100 / 50
s^2 = 2
The hypothesized population variance is σ^2 = 81.
Plugging in the values into the test statistic formula, we get:
t = [(51-1)(2)] / 81
t = 50(2) / 81
t = 100 / 81
t ≈ 1.23
To determine the outcome of the test, we compare the test statistic to the critical value(s) for a 5% significance level. Since the alternative hypothesis is not specific about the direction of the difference, we use a two-tailed test.
Looking up the critical t-value for a two-tailed test with a 5% significance level and 50 degrees of freedom (51-1), we find the critical t-value to be approximately ±2.01.
Since the test statistic (1.23) is less than the critical t-value (-2.01), we do not have enough evidence to reject the null hypothesis. Therefore, we fail to reject H0:µ^2 = 81 at the 5% significance level.
To test the hypothesis H0:µ^2 = 81 vs Ha:µ^2 ≠ 81, we can use a chi-square test for the population variance.
Step 1: Set up the hypotheses:
H0: µ^2 = 81
Ha: µ^2 ≠ 81
Step 2: Determine the significance level:
The significance level is given as 5% or 0.05.
Step 3: Calculate the test statistic:
The test statistic for the chi-square test for population variance is calculated as:
χ^2 = (n - 1) * S^2 / σ^2
Where:
n = sample size
S = sample standard deviation
σ = population standard deviation
Given data:
n = 51
S = 10 (standard deviation)
σ = unknown
We can estimate σ by using S as a substitute.
χ^2 = (51 - 1) * 10^2 / 81
Step 4: Find the critical value for the test statistic:
Since the alternative hypothesis is two-tailed (µ^2 ≠ 81), we split the significance level in half: 0.05 / 2 = 0.025.
With a sample size of 51, the critical value for a chi-square distribution with a significance level of 0.025 and 50 degrees of freedom is approximately 70.222.
Step 5: Make a decision:
If the test statistic χ^2 is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Calculate the outcome of the test:
Now, let's calculate the test statistic χ^2:
χ^2 = (51 - 1) * 10^2 / 81
= 50 * 10^2 / 81
= 5000 / 81
≈ 61.73
Since the test statistic χ^2 (61.73) is greater than the critical value (70.222), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to conclude that the population variance is different from 81 at a 5% significance level.
To perform this hypothesis test, we need to follow several steps. Let's break it down.
Step 1: State the null and alternative hypotheses:
H0 (null hypothesis): The population mean IQ is equal to 81 (µ^2 = 81)
Ha (alternative hypothesis): The population mean IQ is not equal to 81 (µ^2 ≠ 81)
Step 2: Determine the significance level:
The significance level is given as 5%, which means we will reject the null hypothesis if the probability of obtaining our test statistic is less than 5%.
Step 3: Calculate the test statistic:
Since we have the population standard deviation (σ), we can use the Z-test. The formula for the Z-test is:
Z = (X̄ - µ) / (σ / √n)
Where:
X̄ is the sample mean (108),
µ is the hypothesized population mean (81),
σ is the population standard deviation (10),
and n is the sample size (51).
Substituting the given values into the formula, we get:
Z = (108 - 81) / (10 / √51)
= 27 / (10 / √51)
≈ 12.89 (rounded to two decimal places)
Step 4: Determine the critical value(s):
Since the alternative hypothesis is two-tailed (µ^2 ≠ 81), we need to find the critical values corresponding to the 5% significance level.
Using a Z-table or calculator, we find that the critical values are -1.96 and +1.96.
Step 5: Determine the outcome of the test:
Since the absolute value of our test statistic (12.89) is greater than the critical value (1.96), we can reject the null hypothesis. The outcome of the test is that there is sufficient evidence to suggest that the population mean IQ is not equal to 81.
Please note that this explanation assumes that the sample is a random sample from a normally distributed population.