A company sent seven of its employees to attend a course in building self-confidence. These employees were evaluated for their self-confidence before and after attending this course. The following table gives the scores (on a scale of 1 to 15, 1 being the lowest and 15 being the highest score) of these employees before and after they attend the course.
Before 8 5 4 9 6 9 5
After 10 8 5 11 6 7 9
Test at the 1% significance level whether attending this course increases the mean score of employees.
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Use the previous process.
I'll let you do the calculations.
3523
To test whether attending the course increases the mean score of employees, we need to perform a hypothesis test.
The null hypothesis (H0) assumes that attending the course does not increase the mean score of employees. The alternative hypothesis (HA) assumes that attending the course does increase the mean score of employees.
H0: μ(before) = μ(after)
HA: μ(before) < μ(after)
To perform this test, we will use a paired t-test since we have paired observations (before and after scores) for each employee.
Step 1: Calculate the differences between the before and after scores for each employee. We subtract the before score from the after score.
Differences: 2, 3, 1, 2, 0, -2, 4
Step 2: Calculate the mean difference.
Mean difference (D̄): (2 + 3 + 1 + 2 + 0 + (-2) + 4) / 7 = 2/7 ≈ 0.286
Step 3: Calculate the standard deviation of the differences.
Standard deviation of differences (sD): √[(Σ(D - D̄)^2) / (n - 1)]
[(2 - 0.286)^2 + (3 - 0.286)^2 + (1 - 0.286)^2 + (2 - 0.286)^2 + (0 - 0.286)^2 + (-2 - 0.286)^2 + (4 - 0.286)^2] / 6
≈ 3.755
sD ≈ √(3.755) ≈ 1.939
Step 4: Calculate the t-statistic.
t = (D̄ - 0) / (sD / √n)
t = (0.286 - 0) / (1.939 / √7)
≈ 0.286 / 0.731
≈ 0.391
Step 5: Determine the critical value.
Since we are testing whether the mean score increased, we need to look up the critical value for a one-tailed test at the 1% significance level with 6 degrees of freedom (n - 1).
Looking up the critical value in the t-table or using a calculator, we find t_critical ≈ -2.447.
Step 6: Compare the obtained t-value with the critical value.
Since the obtained t-value (0.391) is greater than the critical value (-2.447), we fail to reject the null hypothesis.
Step 7: Interpret the result.
Based on the data, there is not enough evidence to support the claim that attending the course increases the mean score of employees at the 1% significance level.
Therefore, we can conclude that attending the course likely does not have a significant impact on increasing the mean score of employees.