Question

question 1 of 6

1. you want to rent an unfurnished one - bedroom apartment in boston next year. the mean monthly rent for a simple random sample of 32 apartments advertised in the local newspaper is $1,400. assume that the standard deviation is known to be $220. what if the sample size was 40 for the 99% confidence interval. how would the confidence interval change with this larger sample size? (1 point)

the confidence interval would be the same width but shifted to the left
the confidence interval would have the same width but be narrower
the confidence interval would have the same center but be wider
the confidence interval would have the same center but shifted to the right
the confidence interval would be the same width but shifted to the right

Explanation:

Step1: Recall confidence - interval formula

The formula for a confidence interval for the population mean when the population standard - deviation $\sigma$ is known is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size. The width of the confidence interval $W = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$.

Step2: Analyze the effect of sample size on width

As the sample size $n$ increases, the term $\frac{\sigma}{\sqrt{n}}$ decreases. Since $W = 2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ and $z_{\alpha/2}$ and $\sigma$ are constant for a given confidence level and population, when $n$ increases from 32 to 40, the width of the confidence interval $W$ decreases. The center of the confidence interval is the sample mean $\bar{x}$, and increasing the sample size does not change the sample - mean value (assuming the sampling process is unbiased). So the confidence interval will have the same center but be narrower.

Answer:

The confidence interval would have the same center but be narrower