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Statistics / Statistics for Engineers and Scientists 4 / Chapter 5 / Problem 7SE

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

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QUESTION:

Leakage from underground fuel tanks has been a source of water pollution. In a random sample of 87 gasoline stations, 13 were found to have at least one leaking underground tank.

Find a 95% confidence interval for the proportion of gasoline stations with at least one leaking underground tank. How many stations must be sampled so that a 95% confidence interval specifies the proportion to within ±0.03?

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QUESTION:

Leakage from underground fuel tanks has been a source of water pollution. In a random sample of 87 gasoline stations, 13 were found to have at least one leaking underground tank.

Find a 95% confidence interval for the proportion of gasoline stations with at least one leaking underground tank. How many stations must be sampled so that a 95% confidence interval specifies the proportion to within ±0.03?

ANSWER:

Answer:

Step 1 of 2:

(a)

In this question, we are asked to find a 95% confidence interval for the proportion of gasoline stations with at least one leaking underground tank.

In a random sample of 87 gasoline stations, 13 were found to have at least one leaking underground tank.

Let $X$ be the number of success in $n\$ independent Bernoulli trials with success probability $p$ , so that $X\ \sim{}Bin(n,\ p)$ .

Define $\widehat{n}\ =\ n+4,$ and $\widehat{p}\ =\ \frac{X+2}{\widehat{n}}.$ Then a level $100(1\ -\ \alpha{})\%$ confidence interval for p (proportion) is

$\widehat{p\ }\pm{}\ {z}_{\alpha{}/2}\sqrt{\frac{\widehat{p}(1\ -\ \widehat{p})}{\widehat{n}}}$ ………..(1)

Let $X$ represent the gasoline stations with at least one leaking underground tank.

The number of successes is $X\ =\ 13$ , and the number of trials is $n\ =\ 87$ , then we can say $X$ follows the Binomial distribution.

We therefore compute,

$\widehat{n}\ =\ n+4\ =\ 87\ +\ 4\ =\ 91$ ,

$\widehat{p}\ =\ \frac{X+2}{\widehat{n}}.$

= $\frac{13+2}{91}\ =\ 0.1648$

$\sqrt{\frac{\widehat{p}(1\ -\ \widehat{p})}{\widehat{n}}\ }\ =\ \sqrt{\frac{0.1648(1\ -\ 0.1648)}{91}\ }\ \ =\ 0.0389$

For a 95% confidence interval, the value of $\alpha{}/2$ is,

$100(1\ -\ \alpha{})\%\ =\ 95\%$

$(1\ -\ \alpha{})\ =\ 0.95$

$\alpha{}\ =\ 0.05$ and $\alpha{}/2\ =\ 0.025$

Then $z-score\$ for $\alpha{}/2$ is $\pm{}1.64$ , hence $\ {z}_{\alpha{}/2}\ =\ 1.96$

The 95% confidence interval is therefore from equation (1),

$\widehat{p\ }\pm{}\ {z}_{\alpha{}/2}\sqrt{\frac{\widehat{p}(1\ -\ \widehat{p})}{\widehat{n}}}$

$0.1648\pm{}\ 1.96\times{}0.0389\ =\ 0.1648\pm{}0.07622$

(0.0886, 0.241)

Hence 95% confidence interval is (0.0886, 0.241).

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