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 .

Explanation:

Let
`\mu` be the population mean .

As per given , we have

`H_0:\mu=50\\\\ H_a: \mu<50`

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=\frac{\overline{x}-\mu}{(s)/(√(n))}`

Also, it is given that ,

n= 48

`\overline{x}=46`

s= 10

`t=(46-50)/((10)/(√(48)))=(-4)/((10)/(6.93))\\\\=(-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