Question

A researcher wishes to​ estimate, with 90​% ​confidence, the population proportion of families who eat fast food at least once per week. Her estimate must be accurate within ​2% of the population proportion.

​(a) No preliminary estimate is available. Find the minimum sample size needed.
​(b) Find the minimum sample size​ needed, using a prior study that found that 46​% of the respondents said they eat fast food four to six times per week.
​(c) Compare the results from parts​ (a) and​ (b).

Answer 1

To estimate the population proportion with 90% confidence and a 2% margin of error, the minimum sample size with no preliminary estimate is 423. With a preliminary estimate of 46%, the sample size is reduced to 380. Having a prior estimate reduces the required sample size.

Step-by-step explanation:

To calculate the minimum sample size required for the researcher's study, we use the formula for determining sample size in estimating population proportions:

n = (Z²*p*q) / E²

where:

n is the sample size

Z is the z-score corresponding to the desired confidence level

p is the estimated population proportion (or 0.5 when no estimate is available)

q is 1 - p

E is the desired margin of error

Part (a):

With no preliminary estimate, we'll use 0.5 for p for a worst-case scenario as it maximizes the product pq. For a 90% confidence level, the Z-score is approximately 1.645. The desired margin of error E is 0.02.

So, the calculation is:

n = (1.645² * 0.5 * 0.5) / 0.02²

n = 423.

Part (b):

Using the prior estimate of 46%, p = 0.46 and q = 0.54. With the same Z-score and margin of error:

n = (1.645² * 0.46 * 0.54) / 0.02²

n = 380.

Part (c):

The sample size calculated using a preliminary estimate is less than the sample size calculated without one, which shows the importance of having prior information to decrease the required sample size for a given level of precision.