Answer to Question #194478 in Statistics and Probability for Daniela(me)

Question

6. When a random sample of 600 Americans were surveyed, 37% said that their favorite beverage is coffee.

 

a) Find the margin of error as a percent to the nearest tenth. 

 

 

b) If another sample is taken and found that shows 40% reported coffee as their favorite beverage, does this fall in the acceptable range from the first survey?

 

Answer 1

The question is not complete given so i am doing one another question so that you can relate between the two.

Given that the value of 95% confidence is 1.960.

a). A simple random sample of 1000 New Yorkers finds that 87 are left-handed. (a) Find the 95% confidence interval for the proportion of New Yorkers who are left-handed.

$p=\dfrac{87}{1000}=0.087$

$p\pm z \dfrac{\sqrt{p(1-p)}}{n}=0.087+1.960\times \sqrt{\dfrac{0.087(1-0.087)}{1000}}=0.087 \pm0.0175$

We are 95% confident that the proportion of New Yorkers who are left-handed is 0.087 with a margin of error of 0.0175. 

b).Find the 99% confidence interval: 

$p \pm z \sqrt{\dfrac{p(1-p)}{n}}=0.087 7\pm 2.576 \sqrt{\dfrac{0.087(1-0.087)}{1000}}$

$= 0.087 \pm 0.0293=0.0577,0.1163$

We are 99.9% confident that the proportion of New Yorkers who are left-handed is 0.087 with a margin of error of 0.0293.