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

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

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:

`√(V(X)) = √(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 = (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 = √(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 = √(V(X)) = √(np(1-p)) = √(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 = (X - \mu)/(\sigma)`

`Z = (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