Question

A sample standard deviation of 10 weights of packages of grass seed distributed by a certain company was calculated to be 0.286. Assume that weights are normally distributed and find a 95% confidence interval for the standard deviation of all such packages of grass seed

Answer 1

`((9)(0.286)^2)/(19.02) \leq \sigma^2 leq ((9)(0.286)^2)/(2.70)`

`0.0387 \leq \sigma^2 \leq 0.2727`

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

`0.197 \leq \sigma \leq 0.522`

Explanation:

Data given and notation

s=0.286 represent the sample standard deviation

`\bar x` represent the sample mean

n=10 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=10-1=9`

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 table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.025,9)" "=CHISQ.INV(0.975,9)". so for this case the critical values are:

`\chi^2_(\alpha/2)=19.02`

`\chi^2_(1- \alpha/2)=2.70`

And replacing into the formula for the interval we got:

`((9)(0.286)^2)/(19.02) \leq \sigma^2 leq ((9)(0.286)^2)/(2.70)`

`0.0387 \leq \sigma^2 \leq 0.2727`

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

`0.197 \leq \sigma \leq 0.522`