A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 100 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The p-value is

0.0062

Explanation:

p = 1 - P(Z < Zcal)

Zcal = (8300-8000)/ (1200/√100)

Zcal = 2.5

p = 1 - P(Z < Zcal)

= 1 - P(Z < 2.5)

= 1 - 0.9938

= 0.0062

hi : ) i found this so i hope this !

"we first find value of tax in dollars.

the difference between the two prices   is: 23. 54 - 22 = 1.54

therefore the sales tax in dollars is 1.54

the tax % will be   (amount of tax / original cost price ) × 100

(1.54 / 22) × 100  → (0.07) × 100 =   7

so the tax rate is 7%

we can double check the answer.

7 % of 22   → (7/100) × 22 → 0.07 × 22 = 1.54

22 + 1.54 = 23.54"

answer; lags, push the econ below the natural level of output;