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:
[tex]\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[/tex]
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.
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\frac{s}{\sqrt{n}}[/tex]
The critical value of the t is:
[tex]t_{\alpha /2, (n-1)}=t_{0.05/2, (17-1)}=t_{0.025, 16}=2.12[/tex]
*Use a t-table.
Compute the 95% confidence interval for true average as follows:
[tex]CI=438.5294\pm 2.12\cdot\frac{14.988}{\sqrt{17}}[/tex]
[tex]=438.5294\pm 7.7065\\=(430.8229, 446.2359)\\\approx (431, 446)[/tex]
Thus, the 95% confidence interval for true average degree of polymerization is (431, 446).
