Question

A random sample of 20 automobiles has a pollution by-product release standard deviation of 2.3 ounces when 1 gallon of gasoline is used. Find the 90% confidence interval of the population standard deviation. Assume the variable is normally distributed

Answer 1

Test statistic: (n-1)*s^2/sig^2
s is the sample standard deviation given as 2.3 ounces
n is the sample size = 20
sig is the population standard deviation
The test statistic follows a chi-square distribution with 19 degrees of freedom
So, P[x1 <=(n-1)*s^2/sig^2 <= x2] = 0.90
x1 and x2 are the points on a chi-square distribution with 19 df below and above which 5% distribution lies respectively.
x1 = 10.1170
x2 = 30.1435
So, P[sqrt( (n-1)*s^2/x2) <= sig <= sqrt( (n-1)*s^2/x1)] = 0.90

Answer 2

Given: Number of automobiles = 20

Amount of gasoline used = 1 gallon

To find: The 90% confidence interval of the population standard deviation.

Solution:

Let the sample standard deviation which is given by 2.3 ounces be s.

And, the sample size of 20 be n.

The total population standard deviation is given by sig.

According to test statistic, it is known that: $\[\frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}}\]$

∵ The test statistic follows a chi-square distribution with 19 degrees of freedom, it can be said that:

$& P\left[ {{x}_{1}}<=\frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}}<={{x}_{2}} \right]=90 percent \\\\ & \Rightarrow P\left[ {{x}_{1}}<=\frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}}<={{x}_{2}} \right]=\frac{90}{100} \\ & \Rightarrow P\left[ {{x}_{1}}<=\frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}}<={{x}_{2}} \right]=0.90$

$x_{1}$ and $x_{2}$ are two points on any chi-square distribution with 19 degrees below and above where the 5% distribution lies respectively.

Therefore, $x_{1}=10.1170$, and $x_{2}=30.1435$

$\[\therefore P\left[ sqrt\left( \frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}} \right) \right]<=sig<=\left[ sqrt\left( \frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}} \right) \right]=0.90\]$

Hence, the 90% confidence interval of the population standard deviation is $\[\therefore P\left[ sqrt\left( \frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}} \right) \right]<=sig<=\left[ sqrt\left( \frac{(n-1)\times {{s}^{2}}}{si{{g}^{2}}} \right) \right]=0.90\]$.