Question

The following sample of lengths was taken from 8 fluorescent light bulbs off the assembly line. Construct the 95% confidence interval for the population standard deviation for all fluorescent light bulbs tha come off the assembly line. Round your answers to 2 decimal places. 3.4, 3.1, 3.6, 3.3, 2.7, 2.8, 2.4, 3.6Lower endpoint:_________Upper endpoint:_________

Answer 1

`((7)(0.442)^2)/(16.012) \leq \sigma^2 \leq ((7)(0.442)^2)/(1.690)`

`0.0855 \leq \sigma^2 \leq 0.8099`

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

`0.2924 \leq \sigma \leq 0.89995`

Lower endpoint = 0.2924

Upper endpoint= 0.89995

Explanation:

Data given and notation

Data: 3.4, 3.1, 3.6, 3.3, 2.7, 2.8, 2.4, 3.6

We can calculate the sample deviation with the following formula:

`s = \sqrt{(\sum_(i=1)^n (X)i -\bar X)^2)/(n-1)}`

s=0.442 represent the sample standard deviation

`\bar x` represent the sample mean

n=8 the sample size

Confidence=95% or 0.95

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 .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

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

`df=n-1=8-1=7`

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 tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.025,7)" "=CHISQ.INV(0.975,7)". so for this case the critical values are:

`\chi^2_(\alpha/2)=16.012`

`\chi^2_(1- \alpha/2)=1.690`

And replacing into the formula for the interval we got:

`((7)(0.442)^2)/(16.012) \leq \sigma^2 \leq ((7)(0.442)^2)/(1.690)`

`0.0855 \leq \sigma^2 \leq 0.8099`

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

`0.2924 \leq \sigma \leq 0.89995`

Lower endpoint = 0.2924

Upper endpoint= 0.89995