Question

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

Answer 1

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

Step-by-step explanation:

Let $\mu$ be the population mean .

As per given , we have

$H_0:\mu=50\\\\ H_a: \mu$

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 : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

Also, it is given that ,

n= 48

$\overline{x}=46$

s= 10

$t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77$

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value =$t_{\alpha,df}=t_{0.025,47}=2.0117$

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