Inference about standard deviation of normal sample

Question

An experiment was conducted to investigate the filling capability of packing equipment at a winery.

20 bottles were randomly selected, and the fill volume $(in$ $ml)$ was measured.

Assume that the fill volume has a normal distribution. The data is shown below:

\begin{array}{|c|c|c|c|c|} \hline 753& 751 & 752 & 753 & 753 \\ \hline 753& 752& 753& 754& 754\\ \hline 752& 751& 752& 750& 753\\ \hline 755& 753& 756& 751& 750\\ \hline \end{array}

$a)$ Does the data support the claim that the standard deviation of fill volume is less than 1 ml? Use $α=0.05$.

$b)$ Find a 95% two-sided confidence interval on the standard deviation of fill volume. $$\\$$ This is what I have so far:

$a)$

$H_o:σ = 1$
$H_a:σ < 1$

where: $\bar{X}$ = 752.55 and $S_x$ = 1.538112309 $$\\$$

$\chi^2 = \frac{(n-1)s^2}{\sigma^2}$ = $\frac{(19)1.538112309^2}{1^2}$ = 44.950

Is what I did correct?

& I don't know how to find the confidence interval on the standard deviation, I would really appreciate your help!

Answer 1

First, I checked your computation of the sample variance $S^2$ using R statistical software.

x = c(753, 751, 752, 753, 753, 753, 752, 753, 754, 754, 
      752, 751, 752, 750, 753, 755, 753, 756, 751, 750)
var(x)
## 2.365789

(I believe 1.538 is the sample standard deviation; to get a CI, you need to start with the variance.)

For normal data, the quantity $Q = \frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1).$

We use this relationship to find a 95% confidence interval (CI) for $\sigma.$ Because $n = 20$ we use printed tables of $\mathsf{Chisq}(19)$ or software to find lower and upper values, $L = 8.907$ and $U = 32.852$, respectively, that cut 2.5% of the probability from tails of this distribution.

qchisq(c(.025,.975),19)
##  8.906516 32.852327

Thus $$0.95 = P\left(L=8.907 < Q = \frac{19(2.366)}{\sigma^2} = \frac{44.95}{\sigma^2} < U = 32.852\right).$$

After manipulation of the inequalities, we have $$P\left(\frac{44.95}{32.852} < \sigma^2 < \frac{44.95}{8.907}\right) = P(1.368 < \sigma^2 < 5.074) = 0.95.$$

19*var(x)/qchisq(c(.975,.025), 19)
## 1.368244 5.046867

Accordingly, a 95% CI for $\sigma^2$ is $(1.368, 5.074)$ and, upon taking square roots of the endpoints, a 95% Ci for $\sigma$ is $(1.170, 2.246).$ See if you can find a similar formula for the CI of the variance or SD of a normal sample in you text. Or look at the Wikipledia page to which the Answer by @Seyhmus links.

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I am puzzled by part (a). There is no way that a sample SD $1.538 > 1$ provides evidence that $\sigma < 1.$ In fact, this SD provides significant evidence that $\sigma > 1.$ Please check to see if the alternative hypothesis should be $H_a: \sigma \ne 1$ or $H_a: \sigma > 1$ instead. [If $H_a: \sigma < 1$ is correct, then you clearly cannot reject $H_0: \sigma \ge 1.$]

Response to question in comment: Below is Minitab 18 output for testing $H_0: \sigma \ge 1$ vs $H_a: \sigma < 1$ based on $n = 10$ normal observations with $S = 1.538.$

Test and CI for One Variance
Method
σ²: variance of Sample
The chi-square method is valid only for the normal distribution.
Descriptive Statistics
N   StDev   Variance    95% Upper Bound for σ² using Chi-Square
20   1.54       2.37    4.44

Test
Null hypothesis H₀: σ = 1
Alternative hypothesis  H₁: σ < 1
Method  Test
        Statistic   DF  P-Value
Chi-Square  44.95   19  0.999

One would reject $H_0: \sigma = 1$ (or larger) if the P-value were smaller than 5%. But here the P-value is a huge 99.9%. Once again, you have observed $S = 1.538.$ One uses $S$ as an estimate of $\sigma,$ so a value of $S >1$ cannot be taken as evidence that $\sigma < 1.$ Minitab's 'Upper Bound' says that $\sigma^2$ might even be a

Notice that that the sample SD $S = 1.538$ lies within this interval, as it must. However, the sample SD does not lie at the center of this interval. Also, notice that the CI includes values of $\sigma > 1,$ and does not include $\sigma = 1.$ Thus, it is difficult to believe we could have $\sigma < 1.$

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I am puzzled by part (a). There is no way that a sample SD $1.538 > 1$ provides evidence that $\sigma < 1.$ In fact, this SD provides significant evidence that $\sigma > 1.$ Please check to see if the alternative hypothesis should be $H_a: \sigma \ne 1$ or $H_a: \sigma > 1$ instead. [If $H_a: \sigma < 1$ is correct, then you clearly cannot reject $H_0: \sigma \ge 1.$]

Response to question in comment: Below is Minitab 18 output for testing $H_0: \sigma \ge 1$ vs $H_a: \sigma < 1$ based on $n = 10$ normal observations with $S = 1.538.$

Test and CI for One Variance
Method
σ²: variance of Sample
The chi-square method is valid only for the normal distribution.
Descriptive Statistics
N   StDev   Variance    95% Upper Bound for σ² using Chi-Square
20   1.54       2.37    4.44

Test
Null hypothesis H₀: σ = 1
Alternative hypothesis  H₁: σ < 1
Method  Test
        Statistic   DF  P-Value
Chi-Square  44.95   19  0.999

One would reject $H_0: \sigma = 1$ (or larger) if the P-value were smaller than 5%. But here the P-value is a huge 99.9%. Once again, you have observed $S = 1.538.$ One uses $S$ as an estimate of $\sigma,$ so a value of $S >1$ cannot be taken as evidence that $\sigma < 1.$ Minitab's 'Upper Bound' says that $\sigma^2$ might even be a large as 4.44 (equivalent to $\sigma$ as large as 2.11).

By contrast, if $S$ had been something like $S = 0.93,$ then we might wonder whether that is evidence that $\sigma < 1$ or whether such a small $S$ might be due to random error. Then the P-value from Minitab turns out to be 0.372 -- still greater than 0.05, so Yes a $\sigma = 1$ has a good chance of producing such a small $S$ at random.

However, if $S = 0.42,$ then the P-value is much below 0.05, and we can say that an observed $S$ that small is not consistent with $\sigma = 1.$ In that case, we would reject $H_0.$