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Is pos5100 n= by A researcher wishes to estimate with 90% confidence, the population proportion of...

Question

Is pos5100 n= by A researcher wishes to estimate with 90% confidence, the population proportion of...

is pos5100 n= by A researcher wishes to estimate with 90% confidence, the population proportion of adults who support labelin
is pos5100 n= by A researcher wishes to estimate with 90% confidence, the population proportion of adults who support labeling logislation for genetically modified organisms (CMOS) Her estimate must be accurate within 6% of the true proportion io (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 73% of the respondents said they support labeling legislation for GMOs (c) Compare the results from parts (a) and (b). (a) What is the minimum sample size needed assuming that no prior information is available? cribe = (Round up to the nearest whole number as needed) (b) What is the minimum sample size needed using a prior study that found that 73% of the respondents support labeling legislation? TA - (Round up to the nearest whole number as needed) (C) How do the results from (a) and (b) compare? OA. Having an estimate of the population proportion raises the minimum sample size needed OB. Having an estimate of the population proportion reduces the minimum sample size needed OC. Having an estimate of the population proportion has no effect on the minimum sample size neded n xt fo ne fo Click to select your answer(s). matid her hutin rano Tranch - o IL

Answers

Given that, margin of error (E) = 0.06

A 90% confidence level has significance level of 0.10 and critical value is, Z_{\alpha/2} = 1.645

a) We want to find, the sample size, assuming that no prior information is available.

\\ n = \frac {1}{4} *(\frac {Z_{\alpha/2} }{E})^2= \frac {1}{4}*(\frac {1.645}{0.06})^2 = 187.9184 \\\\ => n \approx \bold {188}

=> n = 188

b) We want to find, the sample size for prior estimate = p = 0.73

\\ n = p*(1-p)*(\frac {Z_{\alpha/2} }{E})^2= 0.73*(1-0.73)*(\frac {1.645}{0.06})^2 = 148.1549 \\\\ => n \approx \bold {148}

=> n = 148

c) Here, n = 188 > n = 148

Therefore, having an estimate of the population proportion reduces the minimum sample size needed.

Answer : B)


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