# Thread: Confidence Interval, unknown stand dev

1. ## Confidence Interval, unknown stand dev

...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:

$n = 7$

$1 - {\alpha} = .95$

$\alpha = 0.05$

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

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

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

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

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

$S_X = 6.466$

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

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

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

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

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:
$-0.12 \leq \mu \leq 11.84$

2. I was right the first time. I wasn't sure what you meant by S and S_x....
You left out the square root of n.

the formula is $\bar x \pm t_{n-1,\alpha/2}s/\sqrt{n}$