During a recent drought, a water utility in a certain town sampled 100 residential water bills and found that 73 of the residences had reduced their water consumption over that of the previous year.

a. Find a 95% confidence interval for the proportion of residences that reduced their water consumption.

b. Find a 99% confidence interval for the proportion of residences that reduced their water consumption.

c. Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.05.

d. Find the sample size needed for a 99% confidence interval to specify the proportion to within ±0.05.

e. Someone claims that more than 70% of residences reduced their water consumption. With what level of confidence can this statement be made?

f. If 95% confidence intervals are computed for 200 towns, what is the probability that more than 192 of the confidence intervals cover the true proportions?

Step 1 of 5:

The water utility in a certain town samples 100 residential water bills and found that 73 of the residence are reduced their water consumption over that of previous year.

We have to find

The 95% CI for the proportion of residences that reduce their water consumption.The 99% CI for the proportion of residences that reduce their water consumption.The sample size needed for a 95% CI to specify the proportion within .The sample size needed for a 99% CI to specify the proportion within . At what level of confidence the claim that more than 70% of the residences reduce their water content.The probability that more than 192 of the confidence intervals cover the true proportion.Step 2 of 5:

The 95% CI for the proportion of residences that reduce their water content.

Let X be the number of success in an independent bernoulli trials with success probability P, so that X~B(n,P).

Lets define = n+4.

=

Then a level 100(1-confidence interval for P is

Let X denote the number residences that reduce their water consumption, and n be the total number of residences.

X = 73

n = 100

100+4

= 104

And

= 0.7211

Since here the Z-score is =

We can find from the standard normal table,

So the 95% confidence interval for the proportion of automobiles is

( 0.7211 )

( 0.6349 , 0.8072 )

Therefore the 95% confidence interval for the proportion of residences that reduce their water content is (0.6349, 0.8072).

(b) The 99% CI for the proportion of residences that reduce their water consumption.

Since the Z- score is =

We can find the value from the standard normal table, .

So the 98% CI for the proportion of automobiles whose emission level exceed the standard is

( 0.7211 )

( 0.6076 , 0.8345 )

Therefore the 99% CI for the proportion of residences that reduces their water consumption is (0.6076, 0.8345).