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

Let be the average length of time it took the customers in the sample to check out .

As per given , we have

Since , the alternative hypothesis is right-tailed and population standard deviation is known , so we perform right-tailed t-test.

Also, we have n= 100

Sample mean =

Population standard deviation= s= 0.5

Test statistic :

P-value : P(z>2)=1-P(z<2)=1-0.9772=0.0228

 [using p-value table for z]

Decision : Since p-value (0.0228) < significance level (0.05), so we reject the null hypothesis .

Conclusion : We have enough evidence to accept that  the mean waiting time of all customers is significantly more than 3 minutes.

i.e. At a .05 level of significance, it can be concluded that the mean of the population is more than 3 minutes.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet