Question 362276
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.