Question

the cost of ink cartridges for inkjet 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.
2201.12 ≤ x ≤ 2319.08
(round to two decimal places as needed.)
b. does this sample support the company’s claim? choose the correct answer below.
a. yes, because the company’s claim is between the limits of the confidence interval for the average number of pages yielded by a single black cartridge.
b. no, because the company’s claim is not between the limits of the confidence interval for the average number of pages yielded by a single black cartridge.
c. no, because the company’s claim is between the limits of the confidence interval for the average number of pages yielded by a single black cartridge.
d. yes, because the company’s claim is not between the limits of the confidence interval for the average number of pages yielded by a single black cartridge.

Explanation:

Step1: Recall confidence - interval formula

For a 95% confidence interval of the population mean when the population standard - deviation $\sigma$ is known, 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. The $z$ - score for a 95% confidence interval, $z_{\alpha/2}=1.96$.

Step2: Identify given values

We are given that $\bar{x} = 2260.1$, $\sigma=217$, and $n = 52$.

Step3: Calculate the margin of error

The margin of error $E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=1.96\times\frac{217}{\sqrt{52}}\approx1.96\times\frac{217}{7.2111}\approx1.96\times30.0925\approx58.98$.

Step4: Calculate the confidence interval

The lower limit is $\bar{x}-E=2260.1 - 58.98=2201.12$ and the upper limit is $\bar{x}+E=2260.1 + 58.98=2319.08$. So the 95% confidence interval is $2201.12\leq\mu\leq2319.08$.

Step5: Evaluate the company's claim

The company claims that the average is 2349 pages. Since $2349>2319.08$, the company's claim is not between the limits of the confidence interval.

Answer:

a. $2201.12\leq x\leq2319.08$
b. B. No, because the company's claim is not between the limits of the confidence interval for the average number of pages yielded by a single black cartridge.