Question

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. find a 99% confidence interval for the mean monthly rent for unfurnished one - bedroom apartments available for rent in this community. (1 point) o 1200 & 1600 o 1300 & 1500 o 1300 & 1600 o 1400 & 1500 o 1100 & 1600

Explanation:

Step1: Identify the formula for confidence - interval

For a 99% confidence interval when the population standard - deviation $\sigma$ is known and the sample size $n$ is large ($n\geq30$), the formula is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\sigma$ is the population standard - deviation, $n$ is the sample size, and $z_{\alpha/2}$ is the z - score corresponding to the level of confidence.

Step2: Determine the value of $z_{\alpha/2}$

For a 99% confidence interval, $\alpha = 1 - 0.99=0.01$, so $\alpha/2 = 0.005$. Looking up in the standard normal distribution table, $z_{\alpha/2}=z_{0.005} = 2.576$.

Step3: Identify the given values

We are given that $\bar{x}=1400$, $\sigma = 220$, and $n = 32$.

Step4: Calculate the margin of error $E$

$E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.576\times\frac{220}{\sqrt{32}}\approx2.576\times\frac{220}{5.657}\approx2.576\times38.9=100.2$.

Step5: Calculate the confidence interval

The lower limit is $\bar{x}-E=1400 - 100.2 = 1300$ (rounded to the nearest hundred) and the upper limit is $\bar{x}+E=1400 + 100.2 = 1500$ (rounded to the nearest hundred).

Answer:

1300 & 1500