A random sample of 18 observations is taken from a normal population. The sample mean and sample standard deviation are 76.4 and 4.2, respectively. What is an 80% interval estimate of the population variance?
A) [12.107, 29.735]
B) [10.870, 34.581]
C) [12.819, 31.484]
D) [14.636, 23.443]

Question

Quizplus · Accepted Answer

To estimate the population variance with an 80% confidence interval from a sample, we use the chi-square distribution. The formula involves the sample variance (s^2), the sample size (n), and the chi-square values for the desired confidence level. The sample variance is s^2 = (4.2)^2 = 17.64. With n = 18, degrees of freedom = n - 1 = 17. For an 80% confidence interval, we look up the chi-square values for 0.1 and 0.9 (since it's two-tailed and chi-square distribution is not symmetric) for 17 degrees of freedom. The interval is calculated as [((n-1)s^2)/χ²α/2, ((n-1)s^2)/χ²(1-α/2)]. Plugging in the values and using the appropriate chi-square values (which are not directly provided here but are necessary for the calculation), we get the interval estimate for the population variance as [12.107, 29.735].

A Random Sample of 18 Observations Is Taken from a Normal