Question

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:
418 421 422 422 425 429 431 434 437
439 446 447 449 452 457 461 465

Calculate a two-sided 95% confidence interval for true average degree of polymerization.

Answer 1

Answer:

The 95% confidence interval for true average degree of polymerization is (431, 446).

Step-by-step explanation:

The data provided for the degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range is:

S = {418, 421, 422, 422, 425, 429, 431, 434, 437,  439, 446, 447, 449, 452, 457, 461, 465}

Compute the sample mean and sample standard deviation:

$\bar x=\frac{1}{n}\sum X=\frac{1}{7}\times 7455=438.5294\\\\s=\sqrt{\frac{1}{n-1}\sum (X-\bar x)^{2}}=\sqrt{\frac{1}{17-1}\times 3594.2353}=14.988$

As the population standard deviation is not provided use the t-statistic to compute the two-sided 95% confidence interval for true average degree of polymerization.

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\frac{s}{\sqrt{n}}$

The critical value of the t is:

$t_{\alpha /2, (n-1)}=t_{0.05/2, (17-1)}=t_{0.025, 16}=2.12$

*Use a t-table.

Compute the 95% confidence interval for true average as follows:

$CI=438.5294\pm 2.12\cdot\frac{14.988}{\sqrt{17}}$

     $=438.5294\pm 7.7065\\=(430.8229, 446.2359)\\\approx (431, 446)$

Thus, the 95% confidence interval for true average degree of polymerization is (431, 446).