Question

Urgent, please solve only if you know how to use R. I've attached the data. Thank you!

2) A manufacturing company is interested in whether they can save money by adopting a shorter training period while still achieving desired outcomes for employees. Researchers sampled 15 employees to participate in traditional 3-day training and 12 to participate in revised 2-day training. After the training was complete, the researchers compared exit test scores between the two groups (scores are shown in the following data) score -read.csv(" At alpha 0.05 and assuming that the population is normally distributed, is there evidence that the two methods achieve different results? # If you assume Mu1 and Mu2 are the true mean scores for traditional and revi sed training respectively, then you can set your hypothesis as follows: # H0: Mu1 - Mu2 = 0 #H1: Mu1 - Mu2 not= 0

Data:

traditional.training	revised.training
56	50
50	54
52	45
44	55
52	45
56	60
47	58
53	53
55	45
48	56
42	53
51	52
48	55
49	54
44	56

Answer 1

The two groups are independent of each other, So if X denotes the scores of employees who underwent traditional training and Y denotes the scores of employees who underwent revised training. To check if their means are equal or not, we do a two-sample t test.

After serving the data X and Y as input to R:

We need to type in the following commands, note that here $\alpha=0.05$, we assume that the under lying distribution is normally distributed and homoscedastic.

t.test (x, y ,alternative "two.sided", mu = 0, paired FALSE, var.equal TRUE , conf.level 0.95) Two Sample t-test data: x and y t -1.7784, df 28, p-value 0.0862 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval -6.3120472 0.4453805 sample estimates: mean of x mean of y 49.80000 52.73333

Since, p-value>0.05, we fail to reject the null hypothesis and conclude that there is no evidence to show that the two methods achieve different results.