The manager of a grocery store has taken a random sample of 100 customers. the average length of time it took the customers in the sample to check out was 3.1 minutes. the population standard deviation is known to be 0.5 minute. we want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. at a .05 level of significance, it can be concluded that the mean of the population is

We reject H₀

We have enough evidence to say that the mean of the population is bigger than 3 minutes

Step-by-step explanation:

We have a Normal Distribution problem

1.- Hypothesis:

H₀       null hypothesis                     ⇒   μ₀  =  3

Hₐ   alternative hypothesis             ⇒    μ₀  > 3

2.- Critical value

From z table for a significant level of 0.05

z(c)  = 1.64

z(s)  = ( μ  -  μ₀ ) /( σ/√n)           ⇒  z(s)  = (0.1 * √100)/ 0.5

z(s)  = 1 /0.5

z(s)  = 2

Now we compare z(s)  and z(c)  if  z(s) > z(c)  as it is the case we have a z(s) in the rejection area therefore we reject H₀

42.4

step-by-step explanation:

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