A shopper in a grocery store purchases 22 items on average. A store manager believes that the mean is greater than 22. She surveyed a randomly selected group of 36 customers and found that the mean of the number of items purchased was 23.2. The population standard deviation is 3.7 items. At α = 0.05, can it be concluded that the mean is greater than 22 items?

The is no significant evidence to conclude that mean is greater than 22

Step-by-step explanation:

H0 : μ = 22

H0 : μ > 22

The test statistic :

(xbar - μ) ÷ (σ/sqrt(n))

xbar = 22.3 ; n = 36 ; μ = 22 ; σ = 3.7

(22.3 - 22) ÷ (3.7/√36)

0.3 ÷ 0.6166666

Test statistic = 0.486

The Pvalue using a Pvalue calculator from Zscore is :

Pvalue = 0.313

α = 0.05

Pvalue > α ; We fail to reject the Null ; The is no significant evidence to conclude that mean is greater than 22

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18