Question

A researcher wishes to​ estimate, with 90​% ​confidence, the population proportion of adults who eat fast food four to six times per week. Her estimate must be accurate within 5​% 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 times per week. ​

(c) Compare the results from parts​ (a) and​ (b).

Answer 1

(a) No preliminary estimate is available

The ME = Zcritical * SQRT(p * q/n).

Squaring and solving, n = (Zc/ME)² * p * q

Here ME =5% = 0.05, p = 0.5, q = 0.5

The Zc at = 0.10 is 1.645

Therefore n = (1.645/0.05)² * 0.5 * 0.5 = 270.6

Therefore n = 271 (Rounding to the nearest Integer)

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(b) Data from a prior study is available

Here ME =5% = 0.05, p = 0.4, q = 0.6

The Zc at = 0.10 is 1.645

Therefore n = (1.645/0.05)² * 0.5 * 0.5 = 259.78

Therefore n = 260 (Rounding to the nearest Integer)

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(c) Because the available data proportion is 0.4, which is not too far away from the proportion of 0.5 which was used when data was not unavailable there isn't a great difference in the sample sizes.

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