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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 ?

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Rau7star

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/[tex]\sqrt{n}[/tex] )

z=( 18-15) / (5/[tex]\sqrt{49[/tex] )

z= 3/ 0.714

z= 4.20

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