A teacher gives 200 students a study guide for a test and the average score was 90 with a standard deviation of 6. She did not give the other 200 students a study guide and their average score was 70 with a standard deviation of 8. Find the critical value to determine whether or not the study guide helped students to increase their test score.
What is the significance level you are using?
P = .05? P = .01?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
of the Z score. How does it relate to your P?
To find the critical value for determining whether the study guide helped students increase their test score, we need to perform a hypothesis test. Specifically, we will compare the means of two independent samples.
Let's set up the null and alternative hypotheses:
- Null hypothesis (H0): The study guide did not help students increase their test score. The mean difference between the two groups is equal to 0.
- Alternative hypothesis (Ha): The study guide helped students increase their test score. The mean difference between the two groups is greater than 0.
Next, we need to determine the significance level (alpha) for the test. Assuming a typical level of significance of 0.05, we can use a z-distribution since we have the standard deviations of both samples.
The formula for the test statistic is:
z = (X̄1 - X̄2) / √((σ1^2 / n1) + (σ2^2 / n2))
Where:
X̄1 = mean of the first sample (students who received the study guide)
X̄2 = mean of the second sample (students who did not receive the study guide)
σ1 = standard deviation of the first sample
σ2 = standard deviation of the second sample
n1 = size of the first sample
n2 = size of the second sample
Given:
X̄1 = 90 (mean score of the group that received the study guide)
X̄2 = 70 (mean score of the group that did not receive the study guide)
σ1 = 6 (standard deviation of the group that received the study guide)
σ2 = 8 (standard deviation of the group that did not receive the study guide)
n1 = 200 (size of the group that received the study guide)
n2 = 200 (size of the group that did not receive the study guide)
Now, let's calculate the test statistic:
z = (90 - 70) / √((6^2 / 200) + (8^2 / 200))
Simplifying the equation:
z = 20 / √((6^2 / 200) + (8^2 / 200))
z = 20 / √(0.18 + 0.32)
z = 20 / √0.50
z = 20 / 0.707
z ≈ 28.3
The test statistic (z) calculated is approximately 28.3.
Since we are interested in determining if the study guide helped increase the test score, which implies a one-sided test in this case, we need to find the critical value from the z-distribution for a 95% confidence level.
Looking up the critical value from a standard normal distribution table or using a calculator, we find that the critical value for a one-sided test at a 95% confidence level is approximately 1.645.
Comparing the test statistic (z = 28.3) to the critical value (1.645), we see that the test statistic is significantly larger than the critical value. Therefore, we can reject the null hypothesis.
Conclusion:
The study guide significantly helped students increase their test score, as indicated by the large test statistic and the rejection of the null hypothesis.