Suppose the goal is to estimate the parameter stated in question 3 using a 90% confidence...

Suppose the goal is to estimate the parameter stated in question 3 using a 90% confidence interval with margin of error no larger than 0.039. What is the minimum number of colleges and universities that would need to be selected to allow the calculation of a 90% confidence interval with margin of error no larger than 0.039? It can be assumed for this problem only that the proportion of all colleges and universities that will be moving their summer 2020 classes online is 0.21 and this number can be used for this problem .

Expert Solution

The following information is provided,
Significance Level, α = 0.1, Margin of Error, E = 0.039

The provided estimate of proportion p is, p = 0.21
The critical value for significance level, α = 0.1 is 1.64.

The following formula is used to compute the minimum sample size required to estimate the population proportion p within the required margin of error:
n >= p*(1-p)*(zc/E)^2
n = 0.21*(1 - 0.21)*(1.64/0.039)^2
n = 293.36

Therefore, the sample size needed to satisfy the condition n >= 293.36 and it must be an integer number, we conclude that the minimum required sample size is n = 294
Ans : Sample size, n = 294 or 293

If you take z value upto 3 or 4 deciamal answer would be change

orchestra answered 2 years ago

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