Question

In a survery of 154 households, a Food Marketing Institute found that 106 households spend more than $125 a week on groceries. Please find the 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries.

Answer 1

The 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries is (0.6151, 0.7615).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 154, \pi = (106)/(154) = 0.6883`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.6883 - 1.96\sqrt{(0.6883*0.3117)/(154)} = 0.6151`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6883 + 1.96\sqrt{(0.6883*0.3117)/(154)} = 0.7615`

The 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries is (0.6151, 0.7615).