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 90 of them enjoyed the burger. The sample proportion is 0.409, and the 95% confidence interval for the proportion of residents in Austin who like the new burger is: (0.344, 0.474).1. If you had a random sample of 105 residents instead of 220 the margin of error would? A. Decrease compared to the interval aboveB. Stay the same as compared to the interval aboveC. Increase compared to the interval above2. If you created a 90% confidence interval instead of the 95% confidence interval the margin of error would? A. Decrease compared to the interval aboveB. Stay the same as compared to the interval aboveC. Increase compared to the interval above3. If you created a 99% confidence interval instead of the 95% confidence interval the margin of error would? A. Decrease compared to the interval aboveB. Stay the same as compared to the interval aboveC. Increase compared to the interval above4. If you had a random sample of 370 residents instead of 220 the margin of error would? A. Decrease compared to the interval aboveB. Stay the same as compared to the interval aboveC. Increase compared to the interval above