I really need help with these three questions!

Question

You conduct a survey and find that 422 students out of 813 would vote for Hillary Clinton, if they could vote (same numbers as other problem). What is the likely range for the population proportion?

 

You are conducting a survey and you want to keep your margin of error to 4%. How many participants should you have in your sample?

 

You conduct a survey and find that 422 students out of 813 would vote for Hillary Clinton, if they could vote (same numbers as other problem). What is the margin of error for your survey?

Answer 1

population proportion = 422/813

 

N = 813 

p = 0.519

 

Confidence interval for population proportion

 

Checking assumptions:

 

 

813*0.519 = 422

813*(1-0.519) = 391

 

Both are greater than 10 so this assumption has been met and we can use the standard normal approximation with this data.

 

Now we can compute the standard error for p @ 95%  ; z score = 1.96 

SE(p)=square root [p (1−p)N]=  square root [0.519 *(1-0.519)/813] = 0.0175

 

95% confidence interval for p will be 0.519 + 1.96*0.0175  and 0.519 -  1.96*0.0175

 

Ans : 0.5533 and 0.4847

 

This means that @95% confidence interval, between 55% and 48% of all voters will vote for Hillary Clinton

 

 

If margin of error is 4% - we should use zscore @ 96% confidence interval and repeat the same exercise