Question

# A Food Marketing Institute found that 46% of households spend more than \$125 a week on...

A Food Marketing Institute found that 46% of households spend more than \$125 a week on groceries. Assume the population proportion is 0.46 and a simple random sample of 61 households is selected from the population. What is the probability that the sample proportion of households spending more than \$125 a week is more than than 0.56?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Population proportion  of households spending more than \$125 a week on groceries, P = 46% = 0.46

sample size , n = 61

Variance of proportion = P*(1-P)/n = 0.46*(1-0.46)/61 = 0.46*0.54/61 = 0.2484/61 = 0.004072

Standard deviation of proportion, s = sqrt(Variance) = sqrt(0.004072) = 0.0683

Let, p denotes the sample proportion

Probability that the sample proportion of households spending more than \$125 a week is more than than 0.56 = P[ p > 0.56 ] = P[ ( p - P )/s > ( 0.56 - P )/s ] = P[ ( p - 0.46 )/0.0683 > ( 0.56 - 0.46 )/0.0683 ] = P[ Z > 1.464 ] = 1 - P[ Z < 1.464 ] = 1 - 0.9279 = 0.0721

Probability that the sample proportion of households spending more than \$125 a week is more than than 0.56 = 0.0721

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