Question

The Food Marketing Institute shows that 18% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.18 and a sample of 700 households will be selected from the population. Use z-table.

1. Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals).
     
   
2. What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4 decimals)?
     
   
3. What is the probability that the sample proportion will be within +/- 0.03 of the population proportion for a sample of 1,200 households (to 4 decimals)?
   

Answer 1

Solution :

a)

Given that p = 0.18 , n = 700

=> q = 1-p = 0.82

=> mean µ = p = 0.18

=> standard error σ = sqrt(0.18*0.82/700) = 0.0145

b)

=> Z(0.18 + 0.03) = Z(0.21) = (p^ - p)/sqrt(p*q/n)

= (0.21 - 0.18)/sqrt(0.18*0.82/700)

= 2.0660

=> Z(0.18 - 0.03) = Z(0.15) = (p^ - p)/sqrt(p*q/n)

= (0.15 - 0.18)/sqrt(0.18*0.82/700)

= -2.0660

=> P(0.15 < p < 0.21) = P(-2.0660 < Z < 2.0660) = 0.9616

c)

Given that n = 1200

=> Z(0.18 + 0.03) = Z(0.21) = (p^ - p)/sqrt(p*q/n)

= (0.21 - 0.18)/sqrt(0.18*0.82/1200)

= 2.7050

=> Z(0.18 - 0.03) = Z(0.15) = (p^ - p)/sqrt(p*q/n)

= (0.15 - 0.18)/sqrt(0.18*0.82/1200)

= -2.7050

=> P(0.15 < p < 0.21) = P(-2.7050 < Z < 2.7050) = 0.9932

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