Question

1. Of 300 individuals in a marketing study, 67 reported that they were "Very Satistfied" with the product. Find the 98% confidence interval for this proportion. 2. In one study, 45 executives paid an average of $22 for lunch, with standard deviation $6, and 62 sales associates paid an average of $11, with standard deviation $4. Find the 95% confidence interval for the difference in the mean amounts executives and sales associates spend on lunch. (Do not assume the populations have equal standard deviations. Do show your work.) 3. In a poll of 100 people, 53% favored Candidate A over Candidate B. Is this proportion significantly more than half at the a=5% level? (Show your work.)

Answer 1

1. Here $p=\frac{67}{300}=0.223$

z value for 98% CI is 2.326 as P(-2.326<z<2.326)=0.98

So Margin of Error is $E=z*\sqrt{\frac{pq}{n}}=2.326*\sqrt{\frac{0.223*0.777}{300}}=0.056$

Hence CI is $CI=p \pm E=0.223\pm 0.056=(0.167,0.279)$