Question

Acid rain accumulations in lakes and streams in the northeastern part of the United States are a major environmental concern. A researcher wants to know what fraction of lakes contain hazardous pollution levels. He randomly selects 200 lakes and determines that 45 of the selected lakes have as unsafe concentration of acid rain pollution.

(a) Calculate the best point estimate of the population proportion of lakes that have unsafe concentrations of acid rain pollution.

(b) Determine a 95% confidence interval for the population proportion.

(c) If a local politician states that only 20% of the lakes are contaminated, does the study provide overwhelming evidence at the 95% level to contradict his views?

Answer 1

The best point estimate of the population proportion of lakes with unsafe concentrations of acid rain pollution is 22.5%. The 95% confidence interval for the population proportion is (0.158, 0.292), which does not contain the politician's claim of 20%. Therefore, the study provides overwhelming evidence to contradict the politician's views at the 95% level of confidence.

Step-by-step explanation:

(a) Point Estimate:

The best point estimate of the population proportion of lakes that have unsafe concentrations of acid rain pollution is 45/200 = 0.225 or 22.5%.

Error Bound:

To calculate the error bound, we use the formula:

Error Bound = Z * √(p*(1-p)/n)

Where Z is the z-score corresponding to our desired level of confidence (95%), p is the point estimate, and n is the sample size.

For a 95% confidence interval, the z-score is approximately 1.96. Therefore, the error bound is:

Error Bound = 1.96 * √((0.225*(1-0.225))/200)

Error Bound = 0.067

(b) 95% Confidence Interval:

The 95% confidence interval for the population proportion is given by:

Point Estimate ± Error Bound

0.225 ± 0.067

The confidence interval is (0.158, 0.292).

(c) Contradicting the Politician's Views:

We can use the 95% confidence interval to determine whether the study provides overwhelming evidence to contradict the politician's statement.

If the 95% confidence interval contains the politician's claim (20%), then it does not provide overwhelming evidence to contradict his views. However, if the confidence interval does not contain the politician's claim, then it does provide overwhelming evidence to contradict his views.

In this case, the confidence interval (0.158, 0.292) does not contain the politician's claim of 20%. Therefore, the study does provide overwhelming evidence to contradict the politician's views at the 95% level of confidence.

Answer 2

a) 0.225

b) (0.167, 0.283)

c) No, since 20% = 0.2 is part of the confidence interval.

Step-by-step explanation:

(a) Calculate the best point estimate of the population proportion of lakes that have unsafe concentrations of acid rain pollution.

45 out of 200 lakes. So

45/200 = 0.225

So the best point estimate of the population proportion of lakes that have unsafe concentrations of acid rain pollution is of 0.225.

(b) Determine a 95% confidence interval for the population proportion.

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 200, \pi = 0.225`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.225 - 1.96\sqrt{(0.225*0.775)/(200)} = 0.167`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.225 + 1.96\sqrt{(0.225*0.775)/(200)} = 0.283`

The 95% confidence interval for the population proportion is (0.167, 0.283).

(c) If a local politician states that only 20% of the lakes are contaminated, does the study provide overwhelming evidence at the 95% level to contradict his views?

No, since 20% = 0.2 is part of the confidence interval.