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

This is what i found on google “eightin 1997, the mccaughey septuplets were born in carlisle, iowa. multiple births of as many as eight babies have been born alive, the first set on record to the chukwu family in texas in 1998; one died and seven survived. in 2009, a second set, the suleman octuplets, were born in bellflower, california.” i was surprised about this . i don’t want to have that many babies .