3. A company sent seven of its employees to take a course on how to develop self-confidence. These employees were evaluated before and after taking the course to see what their self-confidence was. The following table presents the scores (on a scale of 1 to 15, where 1 is the lowest score and 15 the highest) of these employees before and after taking the course. Assume that the population of the pair difference has a normal distribution.
Before 8 5 4 9 6 8 5
After 10 7 5 11 6 7 9
a.Test at a 5% significance level if attending the course
increased the average employee score.
b. Build a 99% confidence interval to estimate D.
c. Interpret the conclusion obtained in part a.
1.) be the mean score of the employees before attending the course and be the mean score of the employees after attending the course.
We want to test the claim that attending the course has increased the average employee score. So the hypothesis we frame is,
We perfome a paired-t test to test the given hypothesis and use minitab-16 for the computation purpose.
Steps- Enter the data in different columns> Stat> Basic Statistics> Paired-t> Select the samples( Before score as sample 1)> Under 'options', set the confidence level as 95.0 and alternate as 'less than'> ok
The obtained test statistic is -2.34 and the corresponding p-value is 0.029
As p-value<0.05, we reject the null hypothesis.
2.) To compute 99% confidence interval for the difference-
Steps- Stat> Basic Statistics> Paired-t> select the samples> under 'options',set confidence level as 99.0 and alternate as 'not equal'> ok.
The 99% confidence interval for difference is (-3.696,0.839)
3.) So, we can conclude that at 5% significance level, mean scores after attending the course are better or higher than before.
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