What is the probability that the sample proportion of households spending more than $125 a week is between 0.24 and 0.48?

Question

A Food Marketing Institute found that 27% of households spend more than $125 a week on groceries. Assume the population proportion is 0.27 and a simple random sample of 89 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.24 and 0.48?

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

Answer = (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Hi Paul,

First of all, we need to make sure we have at least ten successes and ten failures. We define "success" as spending over $125/week on groceries. Formula for number of successes is:

np=89*0.27=24.03

Formula for failures is n(1-p), but we can already tell we hit that based on above:

n)1-p)=89*.73=64.97

Now, this means we can assume approximate normality and use a varitation on our classic introductory statistics equation z=(x-mu)/sigma. We have to change this formula because we are working with probabilities and a sample.

z=(p^-p)/sqrt[(p(1-p))/n)]

Breaking this down:

p^=sample proportion

p=population proportion

n=sample size

Now, for this problem, we have two potential values for p^--0.24 and 0.48. We need to compute z-scores for both of them, get the corresponding probabilities from the z-table, and do the subtraction. I will write the z-scores as z₁ and z₂. Same with p^₁ and p^₂.

z₁=(p^₁-p)/sqrt[(p(1-p))/n)]

p^₁=0.24

p=0.27

n=89

z₁=(0.24-0.27)/sqrt((0.27*0.73)/89)

z₁= -0.64 (I cannot round to 4 decimals because z-table does not compute to 4 decimals)

Keep this number in mind; we still need z₂.

z₂=(p^₂-p)/sqrt[(p(1-p))/n)]

p^₂=0.48

p=0.27

n=89

z₂=(0.48-0.27)/sqrt((0.27*0.73)/89)

z₂=4.46

Now, looking at z table, we standardize:

P(Z< -0.64)=p(p<0.24)=0.2611

P(Z<4.46)=p(p<0.48)=1.000--almost a guarantee, statistical software will give more exact value

Now, we do the subtraction:

P=1.000-0.2611

P=0.7389

I hope this helps.