I need help with the part b) of this exercise.
A random sample of $50$ machines obtained that its average life is $\bar{x}=70$ months with a variance of $s^2=49$. Assume that they are normally distributed and based on this:
a) Construct a 90% confidence interval for the population variance.
Degrees of freedom: $n-1=50-1=49$
Critical value:
- Left: $\chi_{0.05,49}^2=33.93031$
qchisq(0.05,49)
[1] 33.93031
- Right: $\chi_{0.95,49}^2=66.33865$
qchisq(0.95,49)
[1] 66.33865
$$(\frac{(50-1)49}{66.33865},\frac{(50-1)49}{33.93031})$$
$$(36.19308,70.7627)$$
b) What sample size will be needed to do a new investigation if we want 96% confidence and a margin of error of 1.5 months?
Here is where I don't know what to do. In my class notes I can't find the formula for a sample size for a confidence interval for a variance. I just find the formula for sample size for a confidence interval for a mean or for a proportion. I both of those formula I see that the squared critical value is multiplied by the variance and that product is divided by the squared margin of error. But how I can find the critical value here if for the confindence interval of a population variance there are two different critical values and I need the degrees of freedom to find them and I don't have the sample size, in fact that's what I'm looking for. :-S
I haven't used here the sample mean. So, I was wondering if this is supposed to be a sample size for a confindence interval of a population mean, but the premise doesn't say it.
HELP! PLEASE!
Thanks in advance for the help.
Let's think about what the second part of the question is asking for. You know that for a $100(1-\alpha)\%$ confidence interval for the variance will be $$\left(\frac{(n-1)s^2}{\chi_{1-\alpha/2,n-1}^2}, \frac{(n-1)s^2}{\chi_{\alpha/2,n-1}^2}\right). \tag{1}$$ So the margin of error is half of the length of this interval, i.e., $$ME = \frac{(n-1)s^2}{2} \left( \frac{1}{\chi_{\alpha/2,n-1}^2} - \frac{1}{\chi_{1-\alpha/2,n-1}^2}\right). \tag{2}$$
We are given $ME = 1.5$, $s^2 = 49$, and $\alpha = 0.04$. Our goal is to find $n$. One way to do this is through guess-and-check. What would be a good initial guess?
I searched online but could not find any formulas to address this question, so I turned to some empirical data analysis. Note that the following is claimed without proof, so any other interested readers may wish to cite the relevant literature, or furnish a proof themselves.
First, I claim that for sufficiently large $n$, $$ME \approx \frac{s^2}{\sqrt{n}} \cdot \sqrt{2} \, z_{1-\alpha/2} \tag{3}$$ where $z_{1-\alpha/2}$ is the $1-\alpha/2$ quantile of the standard normal distribution. Again, I found this formula through empirical study; if anyone has a proof, I would welcome it. Using this, we can easily calculate an initial guess for $n$: we have $$n \approx \frac{2 s^4 z_{1-\alpha/2}^2}{ME^2} \approx 9001.9. \tag{4}$$ Rounding up to the nearest integer $n = 9002$, we substitute this into $(2)$ and get $ME = 1.50074$. Since this is too large, we calculate additional values in the following table:
$$\begin{array}{c|c} n & ME \\ \hline 9003 & 1.50066 \\ 9004 & 1.50057 \\ 9005 & 1.50049 \\ 9006 & 1.50041 \\ 9007 & 1.50032 \\ 9008 & 1.50024 \\ 9009 & 1.50016 \\ 9010 & 1.50007 \\ \color{red}{9011} & \color{red}{1.49999} \\ 9012 & 1.49991 \\ \end{array}$$
The exact value, is $n = 9011$, corresponding to the first value in the table for which $ME \le 1.5$.