An average computer mouse inspector can inspect 50 mice per hour. the 48 computer mice inspectors at a particular factory can only inspect 46 mice per hour with a standard deviation of 10. at a = 0.025, does the company have reason to believe that these inspectors are slower than average?

a. yes, because -2.77 falls in the noncritical region
b. no, because - 1.59 falls in the critical region
c. yes, because -2.77 falls in the critical region
d. no, because -1.59 falls in the noncritical region

 C. Yes, because -2.77 falls in the critical region .

Step-by-step explanation:

Let be the population mean .

As per given , we have

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.

Test statistic :

Also, it is given that ,

n= 48

s= 10

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value =

Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.

So , we reject the null hypothesis.

i.e. -2.77 falls in the critical region.

[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]

Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.

Hence, the correct answer is C. Yes, because -2.77 falls in the critical region

i think that you should ask google for the answer to that problem because if we don't know it google does

yes

step-by-step explanation: