Question

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.

Answer 1

Solution

Given that,

p = 0.17

1 - p = 1 - 0.17 = 0.83

a) n = 700

$\mu$_{$\hat p$} = p = 0.17

$\sigma$_{$\hat p$} =  $\sqrt$[p ( 1 - p ) / n] = $\sqrt$ [(0.17 * 0.83) / 700 ] = 0.0142

b) P( 0.15 < $\hat p$ < 0.19)

= P[(0.15 - 0.17) /0.0142 < ( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (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

$\mu$_{$\hat p$} = p = 0.17

$\sigma$_{$\hat p$} =  $\sqrt$[p ( 1 - p ) / n] = $\sqrt$ [(0.17 * 0.83) / 1400 ] = 0.0100

P( 0.15 < $\hat p$ < 0.19)

= P[(0.15 - 0.17) /0.0100 < ( $\hat p$ - $\mu$ _{$\hat p$} ) / $\sigma$ _{$\hat p$} < (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