Question

he 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 2,392 pages. To test this​ claim, an independent lab measured the page count of 46 cartridges and found the average page count to be 2,269.4. Assume the standard deviation for this population is 216. Complete parts a and b.Create a 95​% confidence interval.

Answer 1

The 95% confidence interval for the average page yield of the black ink cartridge, based on the given sample data and a known population standard deviation, is (2,207.17, 2,331.63).

Step-by-step explanation:

To create a 95% confidence interval for the average page yield of a black ink cartridge, we use sample data and assume that the standard deviation for the population is known. The formula for the confidence interval is:

CI = \( \bar{x} \pm z \times \frac{\sigma}{\sqrt{n}} \)

Where \( \bar{x} \) is the sample mean, \( z \) is the z-score corresponding to the confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

We are given:

• Sample mean (\( \bar{x} \)) = 2,269.4
• Population standard deviation (\( \sigma \)) = 216
• Sample size (n) = 46

The z-score for a 95% confidence level is 1.96. Now we calculate the margin of error:

Margin of Error (ME) = 1.96 \times \frac{216}{\sqrt{46}} \approx 62.2285

The 95% confidence interval is:

CI = 2,269.4 \pm 62.2285

This gives us a range:

Lower Limit = 2,269.4 - 62.2285 = 2,207.17

Upper Limit = 2,269.4 + 62.2285 = 2,331.63

The 95% confidence interval is (2,207.17, 2,331.63).