Question

An assembly operation requires a 1-month training period for a new employee to reach max efficiency. A new method of training was suggested and a test conducted to compare the new method with the standard procedure. Two groups of nine new employees were trained for a period of 3 weeks, one group using the new method and the other following standard training procedure. The length of time in minutes required for each employee to assemble the device was recorded at the end of the 3-week period. The measurements are given in the table.

Operation	Data									Sample Mean	Sample Std. Dev.
Std. Procedure	32,	37,	35,	28,	41,	44,	35,		34	35.22	4.94
New Method	35,	31,	29,	25,	34,	40,	27,		31	31.56	4.48

Assume the population variances are equal. Do the data present sufficient evidence to indicate that the mean time to assemble at the end of the 3-week training period is less for the new training procedure? Follow the steps (a) to (e) using a = .05.

1. What is the null hypothesis?

      H0: μx-μy=0

2. What is the alternative hypothesis?

      H1: μx-μy>0

3. What is the test statistic? Evaluate it.

DF= nx+ny-2=9+9-2=16

sp=4.72

P-Value = P(t > 1.65) = ?

4. What is the rejection region?

?

5. State your conclusion appropriately.

Given that the P-Value is small, it can be concluded that the null hypothesis is rejected and that the mean time to assemble at the end of the 3-week training period is less for the new training procedure?

Answer 1

Let data related to standard procedure is considered as group 1 data

sample mean=m1=35.22 sample SD=S1=4.94 sample size =n1=9

data related to new method is considered as group 2

sample mean=m2=31.56 sample SD=S2=4.48 sample size =n2=9

a)

since we need to test that mean time of group 1 is more than group 2 or not hence null hypothesis and alternative hypothesis are

b)

since we dont know the population variance so use use t statistics

also as we have assumed that both population variances are equal hence we will use pooled standard deviation to calculate the test statistics

now pooled standard deviation is given by

now t statistics is given by

df=n1+n2-2=9+9-2=16

c)

we reject H0 if t statistics > critical value

cirtical value is given by

P(t> critical value)=0.05

from t table with df=16

P(t>1.746)=0.05

so critical value =1.746

hence we reject H0 if t staitistics >critical value

d)

since t statistics (1.64) < critical value (1.746)

Hence we failed to reject H0

hence there is no enough evidence to support the claim that mean time for group1 is more than that of group 2