Question

During a recent drought, a water utility in a certain town sampled 100 residential water bills and found that 72 of the residences had reduced their water consumption over that of the previous year. Find a 95% confidence interval for the proportion of residences that reduced their water consumption. Round the answers to two decimal places. Express your answer as (lower bound, upper bound). The 95% confidence interval is ). Find the sample size needed for a 95% confidence interval to specify the proportion to within +0.04. Round up the answer to the nearest integer. The sample size needed is Someone claims that more than 69.02% of residences reduced their water consumption. With what level of confidence can this statement be made? Express the answer as a percent and round to one decimal place. (Hint: Find the z-score that was used to compute the confidence bound. Note that it is a lower confidence bound. What confidence level is this associated with?) The level of confidence is 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? Round the answer to two decimal places. (Hint: How can we find a probability using the normal approximation to the binomial?) The probability that more than 192 of the confidence intervals cover the true proportions is

Answer 1

To find a 95% confidence interval for the proportion of residences that reduced their water consumption, calculate the sample proportion, the margin of error, and use the formula for the confidence interval. The 95% confidence interval is (0.63, 0.81).

Step-by-step explanation:

To find a 95% confidence interval for the proportion of residences that reduced their water consumption, we can use the formula:

Lower Bound = sample proportion - margin of error

Upper Bound = sample proportion + margin of error

First, calculate the sample proportion:

Sample proportion = number of residences that reduced their water consumption / total sample size

Sample proportion = 72 / 100 = 0.72

Next, calculate the margin of error:

Margin of error = z-score * sqrt[(sample proportion * (1 - sample proportion)) / sample size]

For a 95% confidence level, the z-score is 1.96 (obtained from the z-table).

Now, plug in the values into the formula:

Lower Bound = 0.72 - (1.96 * sqrt[(0.72 * (1 - 0.72)) / 100])

Upper Bound = 0.72 + (1.96 * sqrt[(0.72 * (1 - 0.72)) / 100])

Rounding the answers to two decimal places, the 95% confidence interval for the proportion of residences that reduced their water consumption is (0.63, 0.81).