The widths of 79 randomly selected window blinds were found to have a standard deviation of 2.37. Construct the 95% confidence interval for the population standard deviation of the widths of all window blinds in this factory. Round your answers to two decimal places.

Respuesta :

Answer:

[tex]\frac{(78)(2.37)^2}{104.316} \leq \sigma^2 \leq \frac{(78)(2.37)^2}{55.466}[/tex]

[tex] 4.254 \leq \sigma^2 \leq 8.000[/tex]

And for the deviation we just need to take the square root of the variance and we got:

[tex] 2.06 \leq \sigma^2 \leq 2.83[/tex]

Step-by-step explanation:

Data given and notation

s=2.37 represent the sample standard deviation

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

n=79 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:

[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]

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

[tex]df=n-1=79-1=78[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,78)" "=CHISQ.INV(0.975,78)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=104.316[/tex]

[tex]\chi^2_{1- \alpha/2}=55.466[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(78)(2.37)^2}{104.316} \leq \sigma^2 \leq \frac{(78)(2.37)^2}{55.466}[/tex]

[tex] 4.254 \leq \sigma^2 \leq 8.000[/tex]

And for the deviation we just need to take the square root of the variance and we got:

[tex] 2.06 \leq \sigma^2 \leq 2.83[/tex]