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Soloha48 [4]
3 years ago
13

A random sample of 9 wheels of cheese yielded the following weights in pounds has a sample mean of 20.90 and a sample variance o

f 3.45. Assume the weights of wheels of cheese have a normal distribution. Find a 90% confidence interval for the population variance.
Mathematics
1 answer:
Ronch [10]3 years ago
7 0

Answer:

2.002 \leq \sigma^2 \leq 11.365

Step-by-step explanation:

1) Data given and notation

s represent the sample standard deviation

s^2 represent the sample variance

n=9 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 .

2) 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 \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}^9 (x_i -\bar x)^2}{n-1}}
The sample variance given was [tex]s^2=3.45

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

df=n-1=9-1=8

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 table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,8)" "=CHISQ.INV(0.95,8)". so for this case the critical values are:

\chi^2_{\alpha/2}=15.507

\chi^2_{1- \alpha/2}=2.732

And replacing into the formula for the interval we got:

\frac{(9)(3.45)}{15.507} \leq \sigma \frac{(9)(3.45)}{2.732}

2.002 \leq \sigma^2 \leq 11.365

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