Question

The Food Marketing Institute shows that 16% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.16 and a sample of 600 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 +/- 0.02 of the population proportion (to 4 decimals)?
   
   
3. What is the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,300 households (to 4 decimals)?

Answer 1

a)

Here, p be the true population proportion of households spend more than $100 per week on groceries, n is the sample size.

The formula for the standard error is,

Given

p= 0.16

n = 600

Then the standard error is given by

Therefore,the standard error of the proportion of households spending more than $100 per week on groceries = 0.0150

b)

Therefore,  the probability that the sample proportion will be within +/- 0.02 of the population proportion = 0.8164

c)

Given

p= 0.16

n = 1300

Then the standard error is given by

Now

Therefore,  the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,300 households = 0.9500