The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.17 and a sample of 700 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 within ±0.02 of the population proportion?
(c) Answer part (b) for a sample of 1,400 households.
Solution
Given that,
p = 0.17
1 - p = 1 - 0.17 = 0.83
a) n = 700
= p = 0.17
= [p ( 1 - p ) / n] = [(0.17 * 0.83) / 700 ] = 0.0142
b) P( 0.15 < < 0.19)
= P[(0.15 - 0.17) /0.0142 < ( - ) / < (0.19 - 0.17) /0.0142 ]
= P(-1.41 < z < 1.41)
= P(z < 1.41) - P(z < -1.41)
Using z table,
= 0.9207 - 0.0793
= 0.8414
c) n = 1400
= p = 0.17
= [p ( 1 - p ) / n] = [(0.17 * 0.83) / 1400 ] = 0.0100
P( 0.15 < < 0.19)
= P[(0.15 - 0.17) /0.0100 < ( - ) / < (0.19 - 0.17) /0.0100 ]
= P(-2.00 < z < 2.00)
= P(z < 2.00) - P(z < -2.00)
Using z table,
= 0.9772 - 0.0228
= 0.9544
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