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

Explanation:

The formula for the 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 center of the confidence interval is the sample mean $\bar{x}$, and the width of the confidence interval is $2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. When the sample size $n$ increases (from 32 to 40 in this case), the value of $\frac{\sigma}{\sqrt{n}}$ decreases because $\sigma$ is constant. Since the z - score $z_{\alpha/2}$ for a 99% confidence interval is fixed and $\bar{x}$ (the center of the interval) is not affected by the change in sample size (assuming the sampling process is unbiased), the width of the confidence interval $2z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ will decrease. The center of the confidence interval, which is based on the sample - mean, remains the same as long as the sampling method doesn't introduce a bias in the mean value.

Answer:

The confidence interval would have the same center but be narrower