Question

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. 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

Answer 1

Answer:

20.90% probability that more than 192 of the confidence intervals cover the true proportions

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

A single 95% confidence interval has a 95% probability of covering the true proportions, so $p = 0.95$

200 intervals, so $n = 200$

Then

$\mu = E(X) = np = 200*0.95 = 190$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.95*0.05} = 3.0822$

What is the probability that more than 192 of the confidence intervals cover the true proportions

Using continuity correction, this is $P(X > 192 + 0.5) = P(X > 192.5)$, which is 1 subtracted by the pvalue of Z when X = 192.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{192.5 - 190}{3.0822}$

$Z = 0.81$

$Z = 0.81$ has a pvalue of 0.7910

1 - 0.7910 = 0.2090

20.90% probability that more than 192 of the confidence intervals cover the true proportions