Historically, 10% of the homes in Florida have radon levels higher than recommen

Question

Historically, 10% of the homes in Florida have radon levels higher than recommended by the Environmental Protection Agency. Radon is a weakly radioactive gas know to contribute to health problems. A city in north control Florida has hired an environmental consulting group to determine whether it has a greater than normal problem with this gas. A random sample of 200 homes indicated that 25 had radon levels exceeding EPA recommendations. (a) Construct a 95% confidence interval for the population proportion of homes with excessive levels of radon. (b) Use alpha = 0.05 significance level. Conduct the most appropriate hypothesis test. (c) You would like to write a 95 percent confidence interval for population proportion of homes with excessive levels of radon, with a margin of error equal to E = 0.02. How many homes should you ask to be sampled?

Explanation / Answer

Answer to part a)

Population proportion P = 10% = 0.10

Sample size = 200

x = 25

Sample Proportion p^ = 25/200 = 0.125

Standard error (SE) = sqrt(p*(1-p)/n)

SE = sqrt(0.1 * 0.9 /200)

SE = 0.02121

.

Formula confidence interval is:

p^ - Z*SE , p^ + Z*SE

.

For 95% the value of Z = 1.96

.

On plugging the values in the formula of confidence interval we get:

0.125 - 1.96 *0.02121 , 0.125 + 1.96 * 0.02121

0.125 - 0.04157 , 0.125 + 0.04157

0.08343, 0.16657

.

Answer to part b)

The most appropriate hypothesis can be as follows

Null hypothesis : Population proportion is 10% or 0.10

Ho: P = 0.10

Alternate hypothesis : Population proportion is greater than 10%

Ha: P > 0.10

.

The formula of test statistic is :

Z = (p^ - p) / SE

Z = (0.125 - 0.10) / 0.02121

Z = 1.18

.

On referring to the Z table we get

P(Z < 1.18) = 0.881

Since it is a right tailed test

P value = 1 - 0.881 = 0.119

.

Inference:

Since P value 0.119 is greater than the significance level 0.05

We fail to reject the null hypothesis

.

Conclusion: Thus we conclude that the population Proportion P is 10% or 0.10

.

Answer to part c)

Zc for 95% level is 1.96

E = 0.02

p = 0.1

(1-p) = 0.9

.
formula of sample size is

n = p*(1-p)*(Z/E)^2

n = 0.1 *0.9 * (1.96 /0.02)^2

n = 864.36

Thus the sample size must be 865

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.