Question

According to an independent research, a point estimate of the proportion of U.S. consumers of black tea is p = 0.76. Calculate the sample size needed to be 95% confident that the error in estimating the true value of p is less than 0.015? Use the z-value rounded to two decimal places to obtain the answer. 4072.69

Answer 1

The sample size needed is 3115.

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 z-score that has a p-value of
`1 - (\alpha)/(2)`.

The margin of error is:

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

Point estimate:

`\pi = 0.76`

95% confidence level

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

Calculate the sample size needed to be 95% confident that the error in estimating the true value of p is less than 0.015?

This is n for which M = 0.015. So

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

`0.015 = 1.96\sqrt{(0.76*0.24)/(n)}`

`0.015√(n) = 1.96√(0.76*0.24)`

`√(n) = (1.96√(0.76*0.24))/(0.015)`

`(√(n))^2 = ((1.96√(0.76*0.24))/(0.015))^2`

`n = 3114.26`

Rounding up:

The sample size needed is 3115.