Question

An administrative manager for Universal Engineers, Inc. is studying the frequency of required service on the company’s copy machines. A random sample of six copy machines revealed that the number of copies made before a service technician was called

237000
234011
239194
230795
229000
228000

The manufacturer claims that these copiers typically make 240,000 copies before needing service. At the .05 significance level, is the performance of the copiers different for Universal Engineers from the manufacture claims?

a)State the null hypotheses for this study.

b)State the alternative hypotheses for this study.

c)Select the graph which represents the type of hypothesis test.

d)What is the critical value?   (Enter value with 4 decimal places.

e)Calculate the test statistic.   (Enter value with 3 decimal places

f)Do we reject or not reject the null hypothesis?

g)Provide the p-value

h)What is the definition of the p-value?

I)Based on the computed test statistic or p-value, what is the decision about the average number of copies?

And please sow the equation you use so I can recheck and learn, Thank you.

Answer 1

a) Ho :   µ =   240000
b) Ha :   µ ╪   240000

c) t test

d) critical t value, t* =    ±   2.5706

e)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   4499.9952
Sample Size ,   n =    6
Sample Mean,    x̅ = ΣX/n =    233000.0000
      
degree of freedom=   DF=n-1=   5
      
Standard Error , SE = s/√n =   4499.9952/√6=   1837.1153
t-test statistic= (x̅ - µ )/SE =    (233000-240000)/1837.1153=   -3.810

f) reject the null

g) p-Value   =   0.0125

h) the probability of observing the test statistic extreme or more extreme than observed test statistic , when null hypothesis is true is 0.0125

i) Conclusion: There is enough evidence to conclude that  performance of the copiers is different for Universal Engineers from the manufacture claims