The following sample of lengths was taken from 8 fluorescent light bulbs off the assembly line. Construct the 95% confidence int

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The following sample of lengths was taken from 8 fluorescent light bulbs off the assembly line. Construct the 95% confidence int

erval 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:_________

Mathematics

1 answer:

bazaltina [42] · Answer 1 · 2021-03-17T22:11:39+03:00

Answer:

$\frac{(7)(0.442)^2}{16.012} \leq \sigma^2 \leq \frac{(7)(0.442)^2}{1.690}$

$0.0855 \leq \sigma^2 \leq 0.8099$

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

Step-by-step 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:

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

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:

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

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: