Question

A packaging line fills nominal 64-ounce tomato juice jars with a quantity of juice having anormal distribution with a mean of 64.30 ounces. The process should have a standard deviation smallerthan 0.15 ounces per jar. (A larger standard deviation leads to too many under filled and overfilledjars.) A random sample of 24 jars is taken every hour to evaluate the process. The data is given below under"Data 6". Construct a 95% confidence interval on the process standard deviation. Do the data indicatethat the process standard deviation is greater than 0.15?"Data 6"'ounces'64.3764.2664.2264.4264.1364.4464.6464.1963.8564.1764.2164.2364.6464.1264.9864.3464.264.3164.1564.0964.3364.1964.5764.19

Answer 1

The 95% confidence interval for the variance would be
`0.0321 \leq \sigma^2 \leq 0.1046`.

And taking the square root we got:

`0.1792 \leq \sigma \leq 0.3234`.

Do the data indicatethat the process standard deviation is greater than 0.15?

Yes, since the lower limit of the confidence interval is greater than 0.15 we can conclude that the standard deviation is higher than 0.15 at 5% of significance.

Explanation:

1) Data given and notation

Data: 64.37,64.26,64.22,64.42,64.13,64.44,64.64,64.19,63.85

,64.17,64.21,64.23,64.64,64.12,64.98,64.34,64.2,64.31,64.15

,64.09,64.33,64.19,64.57,64.19

We can calculate the sample standard deviation with the following formula:

`s=\sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

`s^2 =0.05315` represent the sample variance

s=0.2305 represent the sample standard deviation

`\mu =64.3` represent the population mean

`\bar x` represent the sample mea

n=24 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 .

2) 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))`

On this case the sample variance is calculated and for the sample deviation is just the square root of the sample variance.

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

`df=n-1=24-1=23`

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(0.025,23)" "=CHISQ.INV(0.975,23)". so for this case the critical values are:

`\chi^2_(\alpha/2)=38.076`

`\chi^2_(1- \alpha/2)=11.689`

And replacing into the formula for the interval we got:

`((23)(0.05315))/(38.076) \leq \sigma^2 \leq ((23)(0.05315))/(11.689)`

`0.0321 \leq \sigma^2 \leq 0.1046`

So the 95% confidence interval for the variance would be
`0.0321 \leq \sigma^2 \leq 0.1046`.

And taking the square root we got:

`0.1792 \leq \sigma \leq 0.3234`.

Do the data indicatethat the process standard deviation is greater than 0.15?

Yes, since the lower limit of the confidence interval is greater than 0.15 we can conclude that the standard deviation is higher than 0.15 at 5% of significance.