Question

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the flows are generated by waste-water from households and industries, together withsome possible drainage from water stored in the topsoil from previous rainfalls. In a study of an urbansewer system, the following values were obtained for flowrates during dry weather conditions:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Assume that the flow rates are normally distributed.

a. Construct a 98% two-sided confidence interval for the standard deviation of the flow rate under dry weather conditions.
b. Explain, only using words, the meaning of the CI you determined in (a).

Answer 1

a)
`((8)(30.23)^2)/(20.09) \leq \sigma^2 \leq ((8)(30.23)^2)/(1.65)`

`363.90 \leq \sigma^2 \leq 4430.80`

Now we just take square root on both sides of the interval and we got:

`19.08 \leq \sigma \leq 66.56`

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

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 we need to find the sample standard deviation with the following formula:

The sample deviation for this case is s=30.23

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

`df=n-1=9-1=8`

The Confidence interval is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and the critical values are:

`\chi^2_(\alpha/2)=20.09`

`\chi^2_(1- \alpha/2)=1.65`

And replacing into the formula for the interval we got:

`((8)(30.23)^2)/(20.09) \leq \sigma^2 \leq ((8)(30.23)^2)/(1.65)`

`363.90 \leq \sigma^2 \leq 4430.80`

Now we just take square root on both sides of the interval and we got:

`19.08 \leq \sigma \leq 66.56`

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56