The 80% interval estimate of the population variance will be 0.8944.
It is defined as the measure of data disbursement, It gives an idea about how much is the data spread out.
[tex]\rm \sigma = \sqrt{\dfrac{ \sum (x_i-X)}{n}[/tex]
σ is the standard deviation
xi is each value from the data set
X is the mean of the data set
n is the number of observations in the data set.
Given that, 18 observations are chosen at random from a normal population. The mean and standard deviation of the sample are 76.4 and 4.2, respectively.
S² = (x-μ)/n-1
S²=(80-76.4)/18-1
S²=0.8
S=0.8944
Thus, the 80% interval estimate of the population variance will be 0.8944.
Learn more about the standard deviation here:
brainly.com/question/12402189
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