Effect of Outlier on Confidence Interval Width

[needhelp83]

$$\begin{tabular}{lcr} 418 & 421 & 422 & 425& 427 & 431 \tabularnewline 434 & 437 & 439 & 446 & 447 & 448 & 453 \tabularnewline 454 & 463 & 465 \end{tabular}$$

1. Calc the sample mean and sample standard deviation. $$\sum_{i=1}^{17} x_i=7451$$ and $$\sum_{i=1}^{17} x_i^2=3269399$$

n=17 $$\overline{x}=\frac{7451}{17}=438.29$$

s^2=$$\frac{17(\frac{3269399}{17}*(438.29)^2}{16}=\sqrt{233.18}=15.27$$


2. Assuming that the population of degree polymerization has a normal dist, construct a 90% CI for the pop mean degree of polymerization.

$$438.29 \pm T_{.10/2,n-1}*\frac{15.27}{\sqrt{17}}$$
$$438.29 \pm 1.746*\frac{15.27}{\sqrt{17}}$$
$$438.29 \pm 6.466$$
(431.82,444.76)

3. If the value 418 in the original data was 518 instead, would the interval in question 2 be wider or narrower? Explain

I was wondering if I had interpreted this correctly. I answered:

The interval would be (434.14,454.21)

w=width
Since w/2 = $$\alpha / 2 * \sigma / \sqrt{n}$$, we can say that since the sample sd increased the width increased as well.

 

[mighty2000]

Additionally, the sample size remained the same, so the sqrt{n} would not have changed. Therefore, the increase of the sample sd would have increased the width of the interval.