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. (Round your answers to two decimal places.)

Answer 1

Answer:

$438.53-2.12\frac{14.988}{\sqrt{17}}=430.82$    

$438.53+2.12\frac{14.988}{\sqrt{17}}=446.24$    

So on this case the 95% confidence interval would be given by (430.82;446.24)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=438.53$

The sample deviation calculated $s=14.988$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=17-1=16$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,16)".And we see that $t_{\alpha/2}=2.12$

Now we have everything in order to replace into formula (1):

$438.53-2.12\frac{14.988}{\sqrt{17}}=430.82$    

$438.53+2.12\frac{14.988}{\sqrt{17}}=446.24$    

So on this case the 95% confidence interval would be given by (430.82;446.24)