Question

A bottling machine can be regulated so that it discharges an average of µ ounces per bottle. It has been observed that the amount of fill dispensed by the machine is normally distributed with σ = 1 ounce. A sample of n = 9 filled bottles is randomly selected from the output of the machine on a given day (all bottled with the same machine setting) and the ounces of fill measuered for each. 1 i) Find the probability that the sample mean will be within 0.3 ounce of the true mean for that particular setting. [4] ii) How many observations should be included in the sample if we wish X¯ to be within 0.3 ounce of µ with probability 0.95? [4] iii) Suppose we plan to select a random sample of ten bottles and measure the amount of fill in each bottle. If these ten observations are used to calculate S 2 , find numbers b1 and b2 such that P(b1 ≤ S 2 ≤ b2) = 0.90

Answer 1

$a)$

Let $X_1,X_2,X_3,..X_9$ denote the ounces of fill to be observed. Then, $X_i\sim N(\mu,1),\space 1\lt i\lt9$. The sample mean $\bar {X}$ is also normally distributed with mean $\mu$ and variance ${\sigma^2\over n}$. That is, $\bar X\sim N(\mu,{1\over 9})$ as $n=9.$

We need to determine,

$p(|\bar X-\mu|\leq0.3)=p(-0.3\leq\bar X-\mu\leq0.3)$. this can be standardized as follows.

$p(-0.3\leq\bar X-\mu\leq0.3)=p({-0.3\over{1\over\sqrt{9}}}\leq{\bar X-\mu\over {1\over\sqrt{9}}}\leq{0.3\over{1\over\sqrt{9}}})\equiv p({-0.9\lt Z\lt0.9})$

This probability can be written as,

$p({-0.9\lt Z\lt0.9})=\phi(0.9)-\phi(-0.9)=0.8159-0.1841=0.6318$

Therefore, the probability that the sample mean will be within 0.3 ounce of the true mean is 0.6318.

$b)$

Here, we need the sample size $n$ such that, $p(|\bar X-\mu|\leq0.3)=0.95$.This probability can be written as, $p(-0.3\leq\bar X-\mu\leq0.3)=0.95..........(1)$

Dividing each term of the inequality in equation (1) by the standard deviation $\sigma_{\bar x}={1\over\sqrt{n}}$, we have,

$p(-0.3\sqrt{n}\leq{\bar X-\mu\over {1\over\sqrt{n}}}\leq0.3\sqrt{n})=0.95$ but ${\bar x-\mu\over{1\over\sqrt{n}}}=Z$. Thus, $p(-0.3\sqrt{n}\leq{\bar X-\mu\over {1\over\sqrt{n}}}\leq0.3\sqrt{n})=p(-0.3\sqrt{n}\leq Z\leq0.3\sqrt{n})=0.95$ . Using the standard normal tables, the value at $\alpha=0.05$ is $Z_{0.05\over2}=Z_{0.025}=1.96$. Therefore, $p(-0.3\sqrt{n}\leq Z\leq0.3\sqrt{n})=p(-1.96\leq Z\leq1.96)=0.95\implies 0.3\sqrt{n}=1.96$

Thus,

$0.3\sqrt{n}=1.96\implies n=({1.96\over0.3})^2=42.6844\approx 43.$

Therefore, the number of observations which should be included in the sample if we wish $\bar X$ to be within 0.3 ounce of $\mu$ with probability 0.95 is $n=43.$

$c)$

$p(b_1 \leq s^ 2 \leq b_2) = 0.90$

$n=10$

We standardize $s^2$ to a $\chi^2$ random variable as follows.

$p(b_1 \leq s ^2 \leq b_2)=p({(n-1)b_1\over \sigma^2}\leq {(n-1)s^2\over \sigma^2}\leq{ (n-1)b_2\over\sigma^2}) = 0.90$

Since $\sigma^2=1,$ it follows that ${(n-1)s^2\over\sigma^2}=(n-1)s^2$ has a $\chi^2$ distribution with $(n-1)=10-1=9$ degrees of freedom.

Let ${(n-1)b_1\over \sigma^2}=(n-1)b_1$ be $x_1$ and ${(n-1)b_2\over \sigma^2}=(n-1)b_1$ be $x_2$ . We rewrite our probability as, $p(x_1\leq (n-1)s^2\leq x_2)=0.90$

To solve for values of $x_1$ and $x_2$, we proceed as below.

To determine the value of $x_1,$ we find a value that cuts off an area of 0.05 on the lower tail of the $\chi^2$ distribution with 9 degrees of freedom. From the tables, this value is, $\chi^2_{0.05,9}=3.3251$. Thus $x_1=3.3251$. Since $x_1=3.3251\implies (n-1)b_1=3.3251\implies 9b_1=3.3251\implies b_1=0.3695$

Also, to determine the value of $x_2,$ we find a value that cuts off an area of 0.05 on the upper tail of the $\chi^2$ distribution with 9 degrees of freedom. From the tables, this value is, $\chi^2_{0.05,9}=16.9190$. Thus $x_2=16.9190$. Since $x_2=16.9190\implies (n-1)b_2=16.9190\implies 9b_2=3.3251\implies b_2=1.8799$

Therefore, the values of $b_1$ and $b_2$ such that $P(b_1 \leq s^ 2\leq b_2) = 0.90$ are 0.3695 and 1.8799 respectively.