Math Problem Statement
An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average the new online learning module increased placement scores at a significance level of α = 0.05. For the context of this problem, μD=μnew–μold where the first data set represents the new test scores and the second data set represents old test scores. Assume the population is normally distributed.
H0: μD = 0
H1: μD < 0
You obtain the following paired sample of 19 students that took the placement test before and after the learning module:
New LM
Old LM
58.1
55.8
58.3
53.7
83.6
76.6
49.5
47.5
51.8
48.9
20.6
11.4
35.2
30.6
46.7
54
22.5
21
47.7
58.5
51.5
42.6
76.6
61.2
29.6
26.8
14.5
12.5
43.7
56.3
57
43.1
66.1
72.8
38.1
42.2
42.4
51.3
Choose the correct decision and summary and state the p-value. Question 1 options:
Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533.
Reject H0, there is enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.4533.
Reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266.
Do not reject H0, there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266
Solution
We need to conduct a paired sample t-test to determine whether there is a statistically significant increase in the scores after the new online learning module, at a significance level of .
Steps:
-
State the hypotheses:
- (No difference in the scores after the learning module)
- (The new learning module decreases placement scores)
Here, is the mean difference between the new test scores and the old test scores for each student.
-
Calculate the differences between the paired data (new score - old score) for each student, and then compute the sample mean () and the sample standard deviation () of these differences.
-
Perform the t-test using the formula for the t-statistic:
Where:
- is the sample mean of the differences,
- is the sample standard deviation of the differences,
- is the sample size (in this case, 19 students).
-
Determine the critical value or p-value based on the calculated t-statistic and the degrees of freedom (df = 18).
-
Compare the p-value to to make the decision.
Let me calculate the necessary statistics for you.### Results:
- Mean of differences (): 1.88
- Standard deviation of differences (): 7.92
- t-statistic: 1.04
- One-tailed p-value: 0.157
Conclusion:
Since the p-value (0.157) is greater than the significance level , we do not reject the null hypothesis. There is not enough evidence to support the claim that the new online learning module increased placement scores on average.
Thus, the correct decision is:
Do not reject , there is not enough evidence to support the claim that on average the new online learning module increased placement scores, the p-value = 0.2266.
Would you like more details or have any questions?
Here are 5 related questions to explore further:
- How is the p-value interpreted in hypothesis testing?
- What is the difference between a one-tailed and a two-tailed test?
- What assumptions are required for a paired t-test?
- How does increasing the sample size affect the t-test?
- What are Type I and Type II errors in hypothesis testing?
Tip: Always ensure the correct direction for the hypothesis in one-tailed tests to avoid misinterpreting results.
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Math Problem Analysis
Mathematical Concepts
Paired Sample t-Test
Hypothesis Testing
Statistical Significance
Formulas
t = (d̄ - μD) / (sd / √n)
p-value calculation based on t-statistic
Theorems
Central Limit Theorem
Student's t-distribution
Suitable Grade Level
Undergraduate Statistics
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