Question

Two processes in a manufacturing line are performed manually: operation A and operation B. A random sample of 50 different assemblies using operation A shows that the sample average time per assembly is 8.05 minutes, with a population standard deviation of 1.36 minutes. A random sample of 38 different assemblies using operation B shows that the sample average time per assembly is 7.26 minutes, with a population standard deviation of 1.06 minutes.
Required:
For α = 0.10, is there enough evidence in these samples to declare that operation A takes significantly longer to perform than operation B?

Answer 1

Answer:

3.06

Step-by-step explanation:

The correct answer is - 3.06

Reason -

For sample A :

mean, $\bar{x_{1} }$ = 8.05          

Standard deviation, $\sigma_{1}$ = 1.36          

Size, $n_{1}$ = 50          

                 

For sample B :

mean, $\bar{x_{2} }$ = 7.26          

Standard deviation, $\sigma_{2}$ = 1.06          

Size, $n_{2}$ = 38

Now,

$\bar{x_{1} }$ - $\bar{x_{2} }$  = 8.05 - 7.26 = 0.79

Standard deviation error, SE = √($\sigma_{1}$²/$n_{1}$ + $\sigma_{2}$²/$n_{2}$)  = 0.2580

Now,

Z-statistic = ($\bar{x_{1} }$ - $\bar{x_{2} }$ )/ SE

                = 0.79/0.2580

                = 3.06

⇒Z-statistic = 3.06