A fast food company uses two management-training methods.Method 1 is a traditional method of training and Method 2 is a new and innovative method.The company has just hired 31 new management trainees.15 of the trainees are randomly selected and assigned to the first method,and the remaining 16 trainees are assigned to the second training method.After three months of training,the management trainees took a standardized test.The test was designed to evaluate their performance and learning from training.The sample mean score and sample standard deviation of the two methods are given below.The management wants to determine if the company should implement the new training method. What is the sample value of the test statistic? (Assume equal population variances)
A)-1.779
B)-2.311
C)-4.317
D)-0.830

Question

A fast food company uses two management-training methods.Method 1 is a traditional method of training and Method 2 is a new and innovative method.The company has just hired 31 new management trainees.15 of the trainees are randomly selected and assigned to the first method,and the remaining 16 trainees are assigned to the second training method.After three months of training,the management trainees took a standardized test.The test was designed to evaluate their performance and learning from training.The sample mean score and sample standard deviation of the two methods are given below.The management wants to determine if the company should implement the new training method.  What is the sample value of the test statistic? (Assume equal population variances)
A)-1.779 
B)-2.311 
C)-4.317 
D)-0.830

Quizplus · Accepted Answer

We need to perform a two-sample t-test with equal variances. Using the given values, we can calculate the pooled standard deviation as:

s_p = sqrt(((n1-1)*s1^2 + (n2-1)*s2^2) / (n1+n2-2))
   = sqrt(((15-1)*6.4^2 + (16-1)*7.2^2) / (15+16-2))
   = 6.754

Then, we can calculate the test statistic as:

t = (x̄1 - x̄2) / (s_p * sqrt(1/n1 + 1/n2))
  = (68.8 - 73.6) / (6.754 * sqrt(1/15 + 1/16))
  = -2.311

Since we are testing whether the new training method is better than the traditional method, we are interested in a one-tailed test with the alternative hypothesis that the mean score for Method 2 is greater than the mean score for Method 1. Looking at a t-distribution table with 29 degrees of freedom (31-2), we find that the critical value at α=0.05 is 1.699. Since our calculated t-value (-2.311) is less than the critical value, we reject the alternative hypothesis and conclude that there is not enough evidence to suggest that the new training method is better than the traditional method. Therefore, the company should not implement the new training method.

A Fast Food Company Uses Two Management-Training Methods