A clinical psychologist claims that as social support increases this causes life satisfaction to increase.

They want to test this experimentally by having 6 students receiving additional social support in their fall semester and then assessing life satisfaction of participants ( = mu1). Then the same 6 students receive no additional social support in their spring semester and are assessed on their life satisfaction during finals ( = mu2) and they want to assess if the difference in stress with no additional finals relative to additional finals with mu1 - mu2.

= mu1 = Additional social support and

= mu2 = No additional social support

They get the following data:

Additional social support	No Additional social support
86	70
81	47
82	83
78	69
22	53
96	79
What would be the estimated standard error of the differences (i.e., the denominator of your t-statistic)? Please compute and report to 4 decimal places.

To compute the estimated standard error of the differences for this paired sample t-test, we need to follow these steps:

### Step 1: Find the differences between the paired scores
First, calculate the differences between the scores for each participant when they received additional social support and when they did not.

\[
\text{Difference} = \mu_1 - \mu_2
\]

| Additional Social Support (\( \mu_1 \)) | No Additional Social Support (\( \mu_2 \)) | Difference (\( D \)) |
|-----------------------------------------|-------------------------------------------|----------------------|
| 86                                      | 70                                        | \( 86 - 70 = 16 \)    |
| 81                                      | 47                                        | \( 81 - 47 = 34 \)    |
| 82                                      | 83                                        | \( 82 - 83 = -1 \)    |
| 78                                      | 69                                        | \( 78 - 69 = 9 \)     |
| 22                                      | 53                                        | \( 22 - 53 = -31 \)   |
| 96                                      | 79                                        | \( 96 - 79 = 17 \)    |

So the differences are: \( D = [16, 34, -1, 9, -31, 17] \).

### Step 2: Calculate the mean of the differences (\( \overline{D} \))

\[
\overline{D} = \frac{\sum D}{n} = \frac{16 + 34 - 1 + 9 - 31 + 17}{6} = \frac{44}{6} = 7.3333
\]

### Step 3: Calculate the variance of the differences

The variance of the differences is given by:

\[
s_D^2 = \frac{\sum (D - \overline{D})^2}{n - 1}
\]

Now, subtract the mean from each difference and square the result:

\[
(16 - 7.3333)^2 = 75.1111, \quad (34 - 7.3333)^2 = 711.1111, \quad (-1 - 7.3333)^2 = 69.4444
\]
\[
(9 - 7.3333)^2 = 2.7778, \quad (-31 - 7.3333)^2 = 1475.1111, \quad (17 - 7.3333)^2 = 93.4444
\]

Sum these squared differences:

\[
75.1111 + 711.1111 + 69.4444 + 2.7778 + 1475.1111 + 93.4444 = 2426.9999
\]

Now divide by \( n - 1 = 6 - 1 = 5 \):

\[
s_D^2 = \frac{2426.9999}{5} = 485.39998
\]

### Step 4: Calculate the standard deviation of the differences

The standard deviation is the square root of the variance:

\[
s_D = \sqrt{485.39998} = 22.0341
\]

### Step 5: Compute the estimated standard error of the differences (\( SE_D \))

The standard error of the differences is given by:

\[
SE_D = \frac{s_D}{\sqrt{n}} = \frac{22.0341}{\sqrt{6}} = \frac{22.0341}{2.4495} = 9.0000
\]

### Conclusion:
The estimated standard error of the differences is **9.0000** (rounded to four decimal places).

Would you like more details or have any questions?

Here are 5 questions that expand on this information:
1. How is the t-statistic calculated using the standard error of the differences?
2. What assumptions are made in a paired sample t-test?
3. How would you interpret the results if the t-statistic is significant?
4. How does the standard error change with a larger sample size?
5. What are some factors that could affect the life satisfaction scores in this experiment?

**Tip:** Always check that the data meets the assumptions of the paired t-test (normality, independence, etc.) before running the test.

A clinical psychologist claims that as social support increases this causes life satisfaction to increase. They want to test this experimentally by having 6 students receiving additional social support in their fall semester and then assessing life satisfaction of participants (mu1). Then the same 6 students receive no additional social support in their spring semester and are assessed on their life satisfaction during finals (mu2). They want to assess if the difference in stress with no additional finals relative to additional finals with mu1 - mu2. The data collected is as follows: Additional social support = [86, 81, 82, 78, 22, 96] and No additional social support = [70, 47, 83, 69, 53, 79]. What would be the estimated standard error of the differences?

Learn how to calculate the estimated standard error of the differences in a paired sample t-test. This problem uses data on life satisfaction under different levels of social support, exploring key statistical concepts such as mean difference, variance, and standard error. Perfect for undergraduate students studying statistics or psychology.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation