Question

Compute a 95% confidence interval for the population mean, based on the numbers 1, 2, 3, 4, 5, 6, 20. Change the number 20 to 7 and recalculate the confidence interval. Using these
results, describe the effect of an outlier (i.e. extreme value) on confidence interval.

Answer 1

$1.\ s=\sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}}\text{ --- sample variance}.\\ \text{Using Excel (STDEV.S function) we can compute that } s\approx 6.4660.\\ \overline{x}=\frac{41}{7}\\ \text{Critical value } t_{0.05}\approx 2.4469\text{ (we use t-distribution table)}\\ \text{We have:}\\ (41/7-(2.4469)(\frac{6.4660}{\sqrt{7}}),41/7+(2.4469)(\frac{6.4660}{\sqrt{7}}))\\ (-0.1229,11.8372)\\ 2.\ s=\sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}}\text{ --- sample variance}.\\ \text{Using Excel (STDEV.S function) we can compute that } s\approx 2.1602.\\ \overline{x}=4\\ \text{Critical value } t_{0.05}\approx 2.4469\text{ (we use t-distribution table)}\\ \text{We have:}\\ (4-(2.4469)(\frac{2.1602}{\sqrt{7}}),4+(2.4469)(\frac{2.1602}{\sqrt{7}}))\\ (2.0022,5.9978)\\ 3.\ \text{The confidence interval is wider in the first case}\\ \text{because of the outlier 20}.$