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A sample set of weights in pounds are 1.01, 0.95, 1.03, 1.04, 0.97, 0.97, 0.99, 1.01, and 1.03. Assume the population of weights is normally distributed. Find a 99 percent confidence interval for the mean population weight.

Answer :

The 99 percent confidence interval for the mean population weight is (1.002, 1.022).

To find a 99 percent confidence interval for the mean population weight, you can use the following formula:

Confidence Interval = Mean ± (Standard Error * Critical Value)

Where:

  • Mean = (Sum of all the values)/ (Number of values)
  • Standard Error = Standard Deviation / Square Root of (Number of values)
  • Critical Value = The Z-Score for the 99th percentile

In this case, the Mean = (1.01 + .95 + 1.03 + 1.04 + .97 + .97 + .99 + 1.01 + 1.03) / 9 = 1.012

The Standard Deviation = 0.03

The Square Root of 9 is 3

The Z-Score for the 99th percentile is 2.576

Therefore, the 99 percent confidence interval for the mean population weight is 1.012 ± (0.03 / 3 * 2.576) = (1.002, 1.022).

Learn more about confidence interval: https://brainly.com/question/17097944

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