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The mean amount of money spent on lunch per week for a sample of 75 students is $42. If the margin of error for the population mean with a 99% confidence
interval is 2.10, construct a 99% confidence interval for the mean amount of money spent on lunch per week for all students.

Sagot :

Answer:  (39.90, 44.10)

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Explanation:

The confidence interval is centered at the mean of 42. We add and subtract the margin of error (MoE) from this center to get the left and right endpoints of the confidence interval.

  • left endpoint = mean - MoE = 42 - 2.10 = 39.90
  • right endpoint = mean + MoE = 42 + 2.10 = 44.10

The 99% confidence interval is (39.90, 44.10)

This is the same as writing [tex]39.90 < \mu < 44.10[/tex]

We are 99% confident the mean mu is somewhere between 39.90 and 44.10, excluding both endpoints.

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