Question

question 15, 6.3.19-t
part 2 of 3
a researcher wishes to estimate, with 99% confidence, the population proportion of families who eat fast food at least once per week.
her estimate must be accurate within 4% 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 40% of the respondents said they eat fast food four to six
imes per week.
(c) compare the results from parts (a) and (b).
(a) what is the minimum sample size needed assuming that no prior information is available?
n = 1037 (round up to the nearest whole number as needed.)
(b) what is the minimum sample size needed using a prior study that found that 40% of the respondents eat fast food at least once
per week?
n = (round up to the nearest whole number as needed.)

Explanation:

Step1: Recall the formula for sample size

The formula for sample size \(n\) when estimating a proportion is \(n=\frac{z^{2}\cdot p(1 - p)}{E^{2}}\), where \(z\) is the z - score, \(p\) is the proportion, and \(E\) is the margin of error. For a \(99\%\) confidence level, the z - score \(z = 2.576\) (from standard normal distribution tables), and the margin of error \(E=0.04\).

Step2: Substitute the values for part (b)

Given \(p = 0.4\) (from the prior study), we substitute into the formula:

$$
LATEXBLOCK0
$$

Answer:

\(n = 996\) (rounded up to the nearest whole number)