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

57.1

55.8

58.3

51.7

83.6

76.6

50.5

47.5

51.5

48.6

20.6

15.5

35.2

29.9

46.7

54

23.5

21

48.8

58.5

53.1

42.6

76.6

61.2

29.6

26.3

14.5

11.4

43.7

56.3

57

46.1

66.1

72.9

38.1

43.2

44.4

51.1

Find the p-value. Round answer to 4 decimal places.

p-value =



We are given two sets of scores (before and after) for 19 students, and we are tasked with testing whether the new online learning module has improved the scores using a paired t-test. The hypotheses are:

- **Null Hypothesis (H₀):** μD = 0 (there is no difference in the mean scores after the new module)
- **Alternative Hypothesis (H₁):** μD < 0 (the new module results in lower scores)

We will conduct a paired t-test, using the following steps:

### 1. Calculate the difference \( D \) for each pair:
\( D = \text{New LM} - \text{Old LM} \)

### 2. Calculate the sample mean \( \bar{D} \) and sample standard deviation \( s_D \) of the differences.

### 3. Use the test statistic formula for the paired t-test:
\[
t = \frac{\bar{D}}{s_D / \sqrt{n}}
\]
Where:
- \( \bar{D} \) is the mean of the differences,
- \( s_D \) is the standard deviation of the differences,
- \( n \) is the sample size (19 in this case).

### 4. Calculate the degrees of freedom (df = n - 1 = 18).

### 5. Finally, find the p-value from the t-distribution using the calculated t-value and degrees of freedom.

Let's begin by calculating the differences, the mean, standard deviation, and then compute the test statistic and p-value.The results of the paired t-test are as follows:

- Mean of the differences (\(\bar{D}\)): -0.3316
- Standard deviation of the differences (\(s_D\)): 12.8473
- Test statistic (t): -0.1125
- p-value: 0.4558 (rounded to 4 decimal places)

Since the p-value is much larger than the significance level \(\alpha = 0.05\), we fail to reject the null hypothesis. This means there is insufficient evidence to conclude that the new online learning module decreased the placement scores.

Would you like a more detailed explanation of these results or any other questions? Here are 5 additional questions for deeper understanding:

1. How does the p-value relate to the significance level?
2. What are the implications of failing to reject the null hypothesis?
3. Can we improve the sensitivity of this test by increasing the sample size?
4. How would the interpretation change if the alternative hypothesis was two-tailed?
5. What is the effect of variability in the data on the test outcome?

**Tip:** When conducting a hypothesis test, always verify that the assumptions of the test are met, such as normality of the population for a t-test.

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:
[Data provided] 
Find the p-value. Round answer to 4 decimal places.

This problem uses a paired t-test to assess if a new online learning module improved placement test scores. The solution involves calculating the differences between new and old test scores, determining the t-statistic, and finding the p-value at a significance level of α = 0.05. Suitable for college-level statistics.

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