SOLUTION: 41) The food marketing institute shows that 17% of households spend more than $100 per wee on groceries. Assume the population proportion is p=.17 and a simple random sample of 80

Question 362276: 41) The food marketing institute shows that 17% of households spend more than $100 per wee on groceries. Assume the population proportion is p=.17 and a simple random sample of 800 households will be selected from the population.
a) Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries.
b) What is the probability that the sample proportion will be with in plus or minus .02 of the population proportion?
c) Answer part (b) for a sample of 1600.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The food marketing institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p=.17 and a simple random sample of 800 households will be selected from the population.
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a) Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries.
The mean of the sample means = 0.17
The standard deviation of the sample means = sqrt[0.17*0.83/800]
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b) What is the probability that the sample proportion will be within plus or minus .02 of the population proportion?
z(0.19) = (0.19-0.17)/sqrt(0.17*0.83/800] = 1.5060
z(0.15) = -1.5060
P(0.15< p < 0.19) = P(-1.5060< z <1.5060) = 0.8679
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c) Answer part (b) for a sample of 1600.
Put 1600 in place of 800 in the calculations.
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Cheers,
Stan H.

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