...nvrm solved it myself....

Assuming that the population is normally distributed, construct a 95% confidence interval for the population mean, based on the following sample size n = 7: 1, 2, 3, 4, 5, 6, and 20. Change the number 20 to 7 and recalculate the confidence interval. Using these results, describe the effect of an outlier on the confidence interval.

My attempt:

$\displaystyle n = 7$

$\displaystyle 1 - {\alpha} = .95$

$\displaystyle \alpha = 0.05$

$\displaystyle t_{\frac{\alpha}{2}} = t_0.025$

$\displaystyle \mu = \frac{41}{7}$

$\displaystyle S = \sqrt{\frac{1756}{7}} = 15.83847$

$\displaystyle degrees of freedom = n - 1 = 7 - 1 = 6$

$\displaystyle S_X = \frac{S}{\sqrt{n-1}}$

$\displaystyle S_X = 6.466$

$\displaystyle .95 CI = X \pm t_{\frac{\alpha}{2}} S_X$

$\displaystyle .95 CI = (\frac{41}{7} \pm t_{0.025, 6} (6.466)$

$\displaystyle .95 CI = (\frac{41}{7} \pm (2.447) (6.466)$

$\displaystyle .95 CI = (\frac{41}{7} \pm (15.822)$

My answer for the first part of this question is wrong, I don't think I know how to properly take the standard deviation. Please check and correct.

Answer for first part:

$\displaystyle -0.12 \leq \mu \leq 11.84 $