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

`438.53-2.12(14.988)/(√(17))=430.82`

`438.53+2.12(14.988)/(√(17))=446.24`

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

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)(s)/(√(n))` (1)

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

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\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(14.988)/(√(17))=430.82`

`438.53+2.12(14.988)/(√(17))=446.24`

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