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Use the confidence level and sample data to find a confidence interval for estimating the population mu. Round your answer to the same number of decimal places as the sample mean. 37 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 10.3 pounds and a standard deviation of 2.4 pounds. What is the 95% confidence interval for the true mean weight, mu, of all packages received by the parcel service?
a. 9.6 lb < mu < 11.0 lb.
b. 9.4 lb < mu < 11.2 lb.
c. 9.5 lb < mu < 11.1 lb.
d. 9.3 lb< mu < 11.3 lb .

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Answer:

c. 9.5 lb < mu < 11.1 lb.

Step-by-step explanation:

Confidence interval can be stated as M±ME where

  • M is the sample mean (10.3)
  • ME is the margin of error

margin of error (ME) around the mean can be calculated using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

  • z is the corresponding statistic in 95% confidence level (1.96)
  • s is the standard deviation of the sample (2.4)
  • N is the sample size (37)

Putting thesenumbers in the formula we get:

ME=[tex]\frac{1.96*2.4}{\sqrt{37} }[/tex] ≈ 0.7733 ≈ 0.8

Then the 95% confidence interval would be 10.3 ± 0.8

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