Final answer:
To find the minimum sample size needed to estimate the population proportion with a 90% confidence level and an accuracy of within 2%, we can use the formula and substitute the values given in the question. Without a prior estimate, the minimum sample size needed is 427. With a prior estimate of 18%, the minimum sample size needed is 244. Having an estimate of the population proportion reduces the minimum sample size needed.
Step-by-step explanation:
To find the minimum sample size needed to estimate the population proportion with a 90% confidence level and an accuracy of within 2%, we can use the formula:
n = (Zα/2)2 * p(1-p) / E2
Where:
- n is the minimum sample size
- Zα/2 is the critical value corresponding to the desired confidence level (in this case, 90%), which can be obtained from a standard normal distribution table or calculator
- p is the estimated proportion of adults who eat fast food four to six times per week (0.5 if no prior estimate is available)
- E is the desired margin of error (0.02 or 2%)
(a) Plugging in the values, we have:
n = (1.645)2 * 0.5(1-0.5) / 0.022 = 426.42
Since the sample size must be a whole number, we round up to the nearest whole number. Therefore, the minimum sample size needed is 427.
(b) If a prior study found that 18% of the respondents eat fast food four to six times per week, we substitute p = 0.18 into the formula:
n = (1.645)2 * 0.18(1-0.18) / 0.022 = 243.80
Again, rounding up to the nearest whole number, the minimum sample size needed is 244.
(c) Comparing the results from parts (a) and (b), we can see that having an estimate of the population proportion reduces the minimum sample size needed. Therefore, the correct answer is A. Having an estimate of the population proportion reduces the minimum sample size needed.