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ww.matil.com/student/Player Test.aspx?testld 168892812kcenterwinayes Math 2100-08- Fall 2017-Hemphill Test: Final Exam Submit Test This Question: 1 pt 26 of 35 (24 complete) This Test: 35 pts possible A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained i she wishes the estimate to be within 0 03 with 95% confidence if (a) she uses a previous estmate of 0.427 (b) she does not use any prior estimates? (a)na(Round up to the nearest integer) (b) n-L(Round up to the nearest integer) Enter your answer in each of the answer boxes

Explanation / Answer

Solution:

The ESTIMATED SAMPLE PROPORTION should be calculated using the following formula: n = pq*(Zc/E)^2
(a)
Now, with a PRELIMINARY ESTIMATE of 0.42, p = 0.42, and q = 1 - p = 1 - 0.42 = 0.58
Since this is a 95% CONFIDENCE INTERVAL, then significance level is 0.05/2 or 0.025, and so, Zcritical = 1.96
E (Margin of Error) = 0.03
Therefore, n = pq*(Zc/E)^2 becomes:
n = (0.42)(0.58) *(1.96/0.03)^2 = 1039.79307, rounded up to 1040 adults

(b)
Now, since NO PRELIMINARY ESTIMATE WAS GIVEN/WAS AVAILABLE, then 0.5(50%) should be used for p, or for the ASSUMED proportion,
With p being 0.5, and q = 1 - p = 1 - 0.5 = 0.5
Since this is a 95% CONFIDENCE INTERVAL, then significance level is 0.05/2 or 0.025, and so, Zcritical = 1.96
E (Margin of Error) = 0.03
Therefore, n = pq*(Zc/E)^2 becomes:
n = (0.5)(0.5) *(1.96/0.03)^2 = 1067.11111, rounded up to 1068 adults

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