Question

In-N-Out Burger is planning on adding a new burger to the menu. A franchise owner wants to know how well the new burger would sell in Austin, Texas. As such, they want to estimate the proportion of residents in Austin, Texas that would like the new recipe. They randomly select 220 residents and have them taste the burger. Out of these 220 people, they determined that 143 of them enjoyed the burger. The sample proportion is 0.65, and the 95% confidence interval for the proportion of residents in Austin who like the new burger is: (0.587, 0.713).

Required:
a. If you had a random sample of 105 residents instead of 220 the margin of error would:_______
b. If you created a 90% confidence interval instead of the 95% confidence interval the margin of error would:______
c. If you created a 99% confidence interval instead of the 95% confidence interval the margin of error would:_______
d. If you had a random sample of 370 residents instead of 220 the margin of error would:______

Answer 1

90% confidence interval for the population proportion, p? 9.1 ... We will ignore the negative and just use 1.645. ... large a sample is needed in order to be 95% confident and within 3% if ... c) 90% confidence, n=21 ... We cannot use z because we do not have the population standard deviation. ... Find the Margin of Error.