Question

question 7 (1 point)
in developing an interval estimate for a population mean, a sample of 50 observations was used. the interval estimate was 19.76 ± 1.32. had the sample size been 200 instead of 50, the interval estimate would have been
19.76 ± 0.33
9.88 ± 1.32
19.76 ± 0.66
4.94 ± 1.32

Explanation:

Step1: Recall margin - of - error formula

The margin of error $E = z\frac{\sigma}{\sqrt{n}}$. When the population standard deviation $\sigma$ and the z - value (assuming the same confidence level) are constant, the margin of error is inversely proportional to the square root of the sample size $n$, i.e., $E\propto\frac{1}{\sqrt{n}}$.

Step2: Calculate the ratio of sample sizes

The original sample size $n_1 = 50$ and the new sample size $n_2=200$. The ratio $\frac{n_2}{n_1}=\frac{200}{50} = 4$.

Step3: Determine the new margin of error

Since $E\propto\frac{1}{\sqrt{n}}$, if the sample size is multiplied by 4, the margin of error is divided by $\sqrt{4}=2$. The original margin of error $E_1 = 1.32$. The new margin of error $E_2=\frac{E_1}{\sqrt{\frac{n_2}{n_1}}}=\frac{1.32}{\sqrt{4}}=\frac{1.32}{2}=0.66$. The sample mean remains the same. So the new interval estimate is $19.76\pm0.66$.

Answer:

$19.76\pm0.66$