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8.42 the high price of medicines is a source of major expense for those seniors in the united states who have to pay for these medicines themselves. a random sample of 2000 seniors who pay for their medicines showed that they spent an average of $4600 last year on medicines with a standard deviation of $800. make a 95% confidence interval for the corresponding population mean.

Sagot :

The required confidence interval for the corresponding population team is  4564.94  ≤ μ ≤ 4635.06.

Given:

Mean (μ) = $4600

Standard deviation (σ) = $800

Sample size (n) = 2000

Confidence interval = 95%

we have to find the 95% confidence interval for the given population mean. The formula used to find the interval of mean is:

μ ± z [tex]\frac{S.D}{\sqrt{n}}[/tex]

The z value for 95% confidence interval is 1.96.

So, the interval of mean at given confidence level will be,

=  [tex]4600[/tex] ± [tex]1.96(\frac{800}{\sqrt{2000} } )[/tex]

= 4564.94  ≤ μ ≤ 4635.06

Therefore, the required confidence interval for the corresponding population mean is  4564.94  ≤ μ ≤ 4635.06.

To know more about confidence interval here. https://brainly.com/question/13799736#

             

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