Question

A random sample from a normal population is obtained, and the data are given below. Find a 90% confidence interval for . 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545 What is the upper bound of the confidence interval (round off to the nearest integer

Answer 1

Answer:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

$\bar X$ 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

Data given: 114 157 203 257 284 299 305 344 378 410 421 450 478 480 512 533 545

The confidence interval for the population variance $\sigma^2$ is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^{17} (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 362.941$ and the deviation s=132.250

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,16)" "=CHISQ.INV(0.95,16)". so for this case the critical values are:

$\chi^2_{\alpha/2}=26.296$

$\chi^2_{1- \alpha/2}=7.962$

And replacing into the formula for the interval we got:

$\frac{(16)(132.250)^2}{26.296} \leq \sigma \leq \frac{(16)(132.250)^2}{7.962}$

$10641.959 \leq \sigma^2 \leq 35147.074$

Now we just take square root on both sides of the interval and we got:

$103.160 \leq \sigma \leq 187.476$

And the upper bound rounded to the nearest integer would be 187.