Question

the cost of ink cartridges for ink - jet printers can be substantial over the life of a printer. printer manufacturers publish the number of pages that can be printed from an ink cartridge in an effort to attract customers. a company claims that its black ink cartridge will yield an average 2,349 pages. to test this claim, an independent lab measured the page count of 52 cartridges and found the average page count to be 2,260.1. assume the standard deviation for this population is 217. complete parts a and b.
a. create a 95% confidence interval.
□≤x≤□
(round to two decimal places as needed.)

Explanation:

Step1: Identify the formula

For a 95% 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, $z_{\alpha/2}$ is the z - score, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 95% confidence interval, $\alpha = 1 - 0.95=0.05$, so $\alpha/2 = 0.025$, and $z_{\alpha/2}=1.96$.

Step2: Calculate the margin of error $E$

We are given that $\bar{x} = 2260.1$, $\sigma = 217$, and $n = 52$. First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{217}{\sqrt{52}}\approx\frac{217}{7.2111}\approx30.0924$. Then, $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times30.0924\approx58.9811$.

Step3: Calculate the confidence interval

The lower limit is $\bar{x}-E=2260.1 - 58.9811=2201.12$ and the upper limit is $\bar{x}+E=2260.1 + 58.9811=2319.08$.

Answer:

$2201.12\leq x\leq2319.08$