[tex]\begin{tabular}{lcr}(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Small Sample Size

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