## .05 level of significance?!!!

can someone please help me out? i just need the formula. i have no idea how to even start this problem! =( once i can figure out values for X then i can find the average, mean, variance, z score, deviation, etc.

1. The new director of special programs in XYZ Corporation felt the customers were waiting too long to receive and complete forms needed to enroll in special programs. After collecting some data, Ms. Jones determined the mean wait time was 28 minutes. She felt this time period was excessive and she instituted new procedures to streamline the process. One month later, a sample of 127 customers was selected. The mean wait time recorded was 26.9 minutes and the standard deviation of the sampling was 8 minutes.

Using the .05 level of significance, conduct a five-step hypothesis testing procedure to determine if the new processes significantly reduced the wait time.

here's what i've got sofar but i doubt it's correct:

Mean: 26.9 minutes
Stand dev: 8 minutes
.05 of 179 = p < 8.95

Step 1: restate the problem as hypotheses about populations.
Pop 1: Time waited longer than 28 min
Pop 2: Time waited shorter than 28 min
Step 2: determine the characteristics of the comparison distribution, estimate the variance of the mean from sample scores.

S2 = E(x-m)2
N-1

x deviation Sq dev Z score

Sum of sq dev: 164 / (10-1)
Stand dev: 4.049

Step 3: determine the cutoff for rejecting the null hypothesis.
Step 4: figure your sample mean’s score on the comparison distribution/find z score
Step 5: Decide whether or not to reject the null hypothesis(no difference between populations). To accept, the “t” score must be greater than the “t” cut-off sample score.

Researcher will reject the null hypothesis only if the new time waited is in the top 2 1/2 % of the comparison distributuion. +1.96 and -1.96.

Yes we will reject the null hypothesis since the sample’s mean of 28 is not 8.95 or more than the comparison of the new mean, 26.9.