Question

Carl cornfield has been wondering whether he should plant white corn this year. He decides to sample 100 customers. 38% say they would purchase white corn. Carl wants to have 95% confidence interval for this proportion

Answer 1

i think you missed here some details again

Answer 2

for 95% z value is 1.96
Standard error for sample proportion = 0.0485386
\[\sqrt{\frac{ 0.38(1-0.38) }{ 100 }}\]
95% confidence interval
0.38±(1.96∗0.0485386)

interval =
0.38±0.0951

We are 95% confident that the proportion interval proportion of 0.2849 to 0.4751 captures the true population proportion of people who say they will purchase white corn.

Also, p-hat*n > 15 and p-hat(1-p-hat) > 15, s0 we can use the regular proportion standard error formula, because the sample is large enough