Question

A marketing firm claims that its ad campaign will increase sales for a fast-food chain by 15%. In a random sample of 49 stores, the average increase in sales was 18 % with a standard deviation of 5%. What is the z value for the this example rounded to the nearest hundredth ?

Answer 1

Answer:

4.20

Step-by-step explanation:

Given data:

the average increase in sales 'x' = 18

ad campaign will increase sales for a fast-food chain 'μ'= 15

standard deviation 'σ'= 5

random sample of stores 'n'=49

In order to determine z-value, we use the formula i.e

z=( x-μ ) / (σ/$\sqrt{n}$ )

z=( 18-15) / (5/$\sqrt{49$ )

z= 3/ 0.714

z= 4.20