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.

$$mean=\bar{x}=\dfrac{1}{n}\sum_ix_i=\dfrac{1}{7}(1+2+3+4+5$$

$$+6+20)=\dfrac{41}{7}\approx5.857143$$

$$s^2=\dfrac{1}{n-1}\sum_i(x_i-\bar{x})^2=\dfrac{1}{7-1}((1-\dfrac{41}{7})^2+$$

$$+(2-\dfrac{41}{7})^2+(3-\dfrac{41}{7})^2+(4-\dfrac{41}{7})^2$$

$$+(5-\dfrac{41}{7})^2+(6-\dfrac{41}{7})^2+(20-\dfrac{41}{7})^2$$

$$=\dfrac{12292}{294}=\dfrac{1756}{42}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{1756}{42}}\approx6.466028$$

The critical value for $\alpha = 0.05$ and $df = n-1 = 6$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.446899$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5.857143-2.446899\times\dfrac{6.466028}{\sqrt{7}},$$

$$5.857143+2.446899\times\dfrac{6.466028}{\sqrt{7}})$$

$$=(-0.1229, 11.8372)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $-0.1229 < \mu < 11.8372,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(-0.1229, 11.8372).$

2.

$$mean=\bar{x}=\dfrac{1}{n}\sum_ix_i=\dfrac{1}{7}(1+2+3+4+5$$

$$+6+7)=\dfrac{28}{7}=4$$

$$s^2=\dfrac{1}{n-1}\sum_i(x_i-\bar{x})^2=\dfrac{1}{7-1}((1-4)^2+$$

$$+(2-4)^2+(3-4)^2+(4-4)^2$$

$$+(5-4)^2+(6-4)^2+(7-4)^2$$

$$=\dfrac{28}{6}=\dfrac{14}{3}$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{14}{3}}\approx2.160247$$

The critical value for $\alpha = 0.05$ and $df = n-1 = 6$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.446899$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(4-2.446899\times\dfrac{2.160247}{\sqrt{7}},$$

$$4+2.446899\times\dfrac{2.160247}{\sqrt{7}})$$

$$=(2.0021, 5.9979)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $2.0021 < \mu < 5.9979,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(2.0021, 5.9979).$

3.

The confidence interval is wider in the first case because of the outlier 20.