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A Food Marketing Institute found that 53% of households spend more than $125 a week on...

A Food Marketing Institute found that 53% of households spend more than $125 a week on groceries. Assume the population proportion is 0.53 and a simple random sample of 105 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.34?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

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Answer #1

Answer

we have Po = 0.53

sample proportion p 0.34

sample size = 105

We have to use the following formula

P(more than 0.34)= P(z>((hat{p}-p_o)/sqrt{(p_o*(1-p_o))/n}))

putting the given values, we get

P(z > ((0.34ー0.53)/V/(0.53 * (1-0.53))/105)) = P(z > ((-0.19)/00487) )

this gives us

P(z >-3.9008)

using the z distribution, we get

P(z >-3.9008) = 1 (because z greater than -3.901 means z value above 3 standard deviation)

So, required probability is 1.0000 (rounded to four decimal places)

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