The following sample of weights was taken from 8 muffins off the assembly line. Construct the 90% confidence interval for the population variance for all muffins that come off the assembly line. Round your answers to two decimal places. 12.9,12.5,13.0,13.1,12.3,12.9,12.2,13.2

Answer :

Answer:

[tex]\frac{(7)(0.378)^2}{14.067} \leq \sigma^2 \leq \frac{(7)(0.378)^2}{2.167}[/tex]

[tex] 0.0711 \leq \sigma^2 \leq 0.462[/tex]

And rounded we got:

[tex] 0.07 \leq \sigma^2 \leq 0.46[/tex]

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=8 the sample size

Confidence=90% or 0.90

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:

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

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

[tex]s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}

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

[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}[/tex]

The sample mean obtained on this case is [tex]\bar x= 12.7625[/tex] and the deviation s=0.378

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

[tex]df=n-1=8-1=7[/tex]

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

The excel commands would be: "=CHISQ.INV(0.05,7)" "=CHISQ.INV(0.95,7)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=14.067[/tex]

[tex]\chi^2_{1- \alpha/2}=2.167[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(7)(0.378)^2}{14.067} \leq \sigma^2 \leq \frac{(7)(0.378)^2}{2.167}[/tex]

[tex] 0.0711 \leq \sigma^2 \leq 0.462[/tex]

And rounded we got:

[tex] 0.07 \leq \sigma^2 \leq 0.46[/tex]

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