View Single Post
needhelp83
#1
Dec5-10, 07:25 PM
P: 199
[tex]\begin{tabular}{lcr}
418 & 421 & 422 & 425& 427 & 431 \tabularnewline
434 & 437 & 439 & 446 & 447 & 448 & 453 \tabularnewline
454 & 463 & 465
\end{tabular}[/tex]

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

n=17 [tex]\overline{x}=\frac{7451}{17}=438.29[/tex]

s^2=[tex]\frac{17(\frac{3269399}{17}*(438.29)^2}{16}=\sqrt{233.18}=15.27[/tex]


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

[tex]438.29 \pm T_{.10/2,n-1}*\frac{15.27}{\sqrt{17}}[/tex]
[tex]438.29 \pm 1.746*\frac{15.27}{\sqrt{17}}[/tex]
[tex]438.29 \pm 6.466[/tex]
(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 = [tex]\alpha / 2 * \sigma / \sqrt{n}[/tex], we can say that since the sample sd increased the width increased as well.
Phys.Org News Partner Science news on Phys.org
Bees able to spot which flowers offer best rewards before landing
Classic Lewis Carroll character inspires new ecological model
When cooperation counts: Researchers find sperm benefit from grouping together in mice