On Clinic
Statistics?
From a random sample of 2000 internet users was carried out to test who whould prefer to find out medical info. online or from a library. There were 1400 responses of which 872 preferred to use the internet. A) what porportion p-bar would use the internet? B) what is the standard error of the poportion p_bar that would use the internet instead of the library? C) how large a sample size would be required for a margin of error of at most .01 in a 99% Confidence level to estimate the porportion pbar that prefer the internet instead of the library?
Public Comments
- A) 872/1400 = 0.623 B) The standard error is s/√n, where s = √(np(1-p)) = 18.13 So the standard error is: 0.485 C) Let margin of error, E = 0.01, and we can use p = 0.623 Then, n = [Zα/2/E]²p(1-p), Where Zα/2 = 2.575 for a 99% confidence interval. n = [2.575/0.010]²(0.623)(1-0.623) ≈ 15,574
- a) 872 / 1400 b) the standard error = sqrt( pbar * (1 - pbar) / n) where n is the sample size. = sqrt( (872 / 1400) * (1 - 872 / 1400) / 1400) = 0.01295338 c) large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval. For large sample confidence intervals for the proportion in this situation you have: pHat ± z * sqrt( (pHat * (1-pHat)) / n) where pHat is the sample proportion z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α n is the sample size in this case you have a z-score such that: P( |Z| < z) = 0.99 P( |Z| > z) = 0.01 P( Z > z) = 0.005 P( Z > 2.575829) = 0.005 to restrict the margin or error we have: z * sqrt( (pHat * (1-pHat)) / n) < 0.01 2.575829 * sqrt( (pHat * (1-pHat)) / n) < 0.01 2.575829 * sqrt( ( (872/1400) * (1- (872/1400))) / n) < 0.01 we use the proportion from the experiment as a starting point and solve for n 2.575829 * sqrt( ( (872/1400) * (1- (872/1400))) / n) < 0.01 2.575829 * sqrt( 0.2349061 / n ) < 0.01 sqrt( 0.2349061 / n ) < 0.01 / 2.575829 0.2349061 / n < 1.507183e-05 n > 0.2349061 / 1.507183e-05 n > 15585.77 n must be an integer value so n > 15586
