Question

1. in a survey of 930 students in a school, 92% reported having pets at home. what is the margin of error? use the formula $pm\frac{1}{sqrt{n}}$ where n is the population. (4 points)
2. in a survey of 850 students in a school, 90% reported having pets at home. if the margin of error is $pm3.4%$, what is the interval that is likely to contain the exact percent of all people who have pets at home? (4 points)

Explanation:

Step1: Identify the value of n for the first - question

The number of students surveyed $n = 930$.

Step2: Calculate the margin of error

Use the formula $\pm\frac{1}{\sqrt{n}}$. Substitute $n = 930$ into the formula: $\pm\frac{1}{\sqrt{930}}\approx\pm\frac{1}{30.496}\approx\pm 0.033 = \pm 3.3\%$.

Step3: For the second - question

The sample percentage is $p = 90\%$ and the margin of error $E=\pm 3.4\%$.

Step4: Calculate the confidence interval

The lower limit of the interval is $p - E=90\% - 3.4\% = 86.6\%$.
The upper limit of the interval is $p + E=90\%+3.4\% = 93.4\%$.

Answer:

1. The margin of error for the first survey is approximately $\pm 3.3\%$.
2. The interval that is likely to contain the exact percent of all people who have pets at home for the second survey is $86.6\%$ to $93.4\%$.