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

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

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=(1)/(n)\sum X=(1)/(7)* 7455=438.5294\\\\s=\sqrt{(1)/(n-1)\sum (X-\bar x)^(2)}=\sqrt{(1)/(17-1)* 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(s)/(√(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(14.988)/(√(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).