A national grocer’s magazine reports the typical shopper spends 11.5 minutes in line waiting to check out. A sample of 19 shoppers at the local Farmer Jack’s showed a mean of 10.7 minutes with a standard deviation of 3.1 minutes. Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the 0.050 significance level.

There is no significant evidence to support that waiting time is less than the mean reported.

Step-by-step explanation:

Given that :

Sample mean, x = 10.7

Standard deviation, s = 3.1

Sample, size , n = 19

H0: μ ≥ 11.5

H1: μ < 11.5

Using the t-test (sample size is small)

Test statistic :

(x - μ) / s/sqrt(n)

(10.7 - 11.5) / (3.1 / sqrt(19))

-0.8 / 0.7111887

= −1.124877

Tcritical = Tdf, 0.05

df = n - 1 ; T18,0.05/2 = - 2.101 (left tailed)

Decision region:

Reject H0 :

If Tstatistic < Tcritical

−1.124877 > - 2.101

Hence, We fail to reject H0

There is no significant evidence to support that waiting time is less than the mean reported.

answer: sensy

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b