Asimple random sample of 10 observations was taken from a large normal population. the sample mean and the population standard deviation were determined to be 80 and 12, respectively. a 99% population mean confidence interval is approximately

What we are looking for is the confidence interval.

Given:

Sample it is denoted by n = 10 

Mean = 80

Standard deviation = 12

Confidence Interval = 99%

Now compute for the margin of error, the formula is Za/2 * σ/√(n)

A 99% level of confidence has a critical value of zα/2 = 2.58.

The standard error is σ/√(n) = 3.162

Multiply 2.58 by 3.162 = 9.8 is the margin of error.

The formula for confidence interval is x̅ ± Za/2 * σ/√(n)

= 80 ± 9.8

= 70.2 < x< 89.8

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