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 600 households will be selected from the population. Use z-table.

A) Calculate the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals).

I got .0153

B) What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4 decimals)?

The correct answer for part B is .94.96 but I do not know how to get this answer?

Answer 1

a )

Standard error = sqrt [ p ( 1 - p) / n ]

= sqrt [ 0.17 ( 1- 0.17) / 600 ]

= 0.01534

b)

Using normal approximation,

P( < p) = P( Z < ( - p) / SE )

So

P( p - 0.03 < P < p + 0.03) = P(0.14 < p < 0.20)

= P( < 0.20) - P( < 0.14)

= P(Z < (0.20 - 0.17) / 0.01534 ) -  P(Z < (0.14 - 0.17) / 0.01534 )

= P(Z < 1.956) - P(Z < -1.956)

P(Z < 1.956) - ( 1 - P(Z < 1.956)

= 0.9748 - ( 1 - 0.9748)

= 0.9496