Question

a researcher wishes to estimate, with 92% confidence, the population proportion of likely u.s. voters who think congress is doing a good or excellent job. her estimate must be accurate within 2% of the true 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 22% of the respondents said they think congress is doing a good or excellent job.
(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 = ) (round up to the nearest whole number as needed.)
(b) what is the minimum sample size needed using a prior study that found that 22% of the respondents said they think congress is doing a good or excellent job?
( n = ) (round up to the nearest whole number as needed.)
(c) how do the results from (a) and (b) compare?
a. having an estimate of the population proportion has no effect on the minimum sample size needed.
b. having an estimate of the population proportion reduces the minimum sample size needed.
c. having an estimate of the population proportion increases the minimum sample size needed.

Explanation:

Step1: Find \(z_{\alpha/2}\) for 92% confidence level

For 92% confidence level, \(\alpha = 1 - 0.92=0.08\), \(\alpha/2 = 0.04\). Using the standard normal distribution \(z_{\alpha/2}=z_{0.04}\approx1.75\) (from standard normal table). The formula for sample size \(n=\frac{z_{\alpha/2}^{2}\cdot p(1 - p)}{E^{2}}\), where \(E = 0.02\) (margin of error).

Part (a)

Step1: Use \(p = 0.5\) (when no prior estimate)

Substitute \(z_{\alpha/2}=1.75\), \(p = 0.5\), \(1 - p=0.5\), \(E = 0.02\) into \(n=\frac{z_{\alpha/2}^{2}\cdot p(1 - p)}{E^{2}}\)

$$n=\frac{(1.75)^{2}\times0.5\times0.5}{(0.02)^{2}}=\frac{3.0625\times0.25}{0.0004}=\frac{0.765625}{0.0004}=1914.0625$$

Step2: Round up

Since \(n\) must be an integer, \(n = 1915\)

Part (b)

Step1: Use \(p = 0.22\), \(1 - p=0.78\)

Substitute \(z_{\alpha/2}=1.75\), \(p = 0.22\), \(1 - p = 0.78\), \(E = 0.02\) into \(n=\frac{z_{\alpha/2}^{2}\cdot p(1 - p)}{E^{2}}\)

$$n=\frac{(1.75)^{2}\times0.22\times0.78}{(0.02)^{2}}=\frac{3.0625\times0.1716}{0.0004}=\frac{0.52545}{0.0004}=1313.625$$

Step2: Round up

Since \(n\) must be an integer, \(n = 1314\)

Part (c)

Step1: Compare the two sample - sizes

We have \(n_a=1915\) (from part (a)) and \(n_b = 1314\) (from part (b)).

Answer:

(a) \(n = 1915\)
(b) \(n = 1314\)
(c) Assuming an estimate of the population proportion reduces the minimum sample size needed.