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Chapter 6: Problem 48

Compilation of large masses of data on lung cancer shows that approximately 1 of every 40 adults acquires the disease. Workers in a certain occupation are known to work in an air-polluted environment that may cause an increased rate of lung cancer. A random sample of \(n=400\) workers shows 19 with identifiable cases of lung cancer. Do the data provide sufficient evidence to indicate a higher rate of lung cancer for these workers than for the national average?

Short Answer

Expert verified
Answer: Yes, we can conclude that the rate of lung cancer among the workers in this occupation is higher than the national average, as the hypothesis test provided sufficient evidence to reject the null hypothesis.

Step by step solution

01

Define the null and alternative hypotheses

Let p be the national proportion of adults with lung cancer and p_work be the proportion of workers in the given occupation with lung cancer. Null hypothesis (H0): p_work = p Alternative hypothesis (Ha): p_work > p
02

Identify the test statistic and its distribution

We will use the Z-test to compare the sample proportion with the population proportion. If the null hypothesis is true, the test statistic follows a standard normal distribution, i.e., N(0,1).
03

Calculate the sample proportion and the test statistic

First, calculate the sample proportion by dividing the number of workers with lung cancer by the total number of workers in the sample. p_hat = 19/400 = 0.0475 Now, calculate the test statistic Z using the following formula: Z = (p_hat - p) / sqrt(p * (1-p) / n) where p is the national average proportion of adults with lung cancer, which is 1/40 = 0.025, and n is the sample size, which is 400 workers. Z = (0.0475 - 0.025) / sqrt(0.025 * (1-0.025) / 400) Z = 0.0225 / sqrt(0.000151875) Z 鈮� 1.83
04

Determine a significance level

We will use a significance level of 伪 = 0.05, which is commonly used in hypothesis testing.
05

Find the critical value and make a decision

Since the alternative hypothesis is right-tailed (Ha: p_work > p), we will find the critical value for a right-tailed test. Using the Z-table or a calculator, we can determine the critical value for a 0.05 significance level: Z_critical = 1.645 Now, compare the calculated test statistic, Z, with the critical value, Z_critical: Since 1.83 > 1.645, we reject the null hypothesis (H0).
06

Conclusion

Since we rejected the null hypothesis, we have sufficient evidence to conclude that the rate of lung cancer among the workers in this occupation is higher than the national average.

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Most popular questions from this chapter

Consider a binomial random varible with \(n=25\) and \(p=.6 .\) Fill in the blanks below to find some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate \(n p=\) _____ and \(n q=\) _____ b. Are \(n p\) and \(n q\) both greater than \(5 ?\) Yes ____ No ____ c. If the answer to part \(b\) is yes, calculate \(\mu=n p=\) ______ and \(\sigma=\sqrt{n p q}=\) ______ d. To find the probability of more than 9 successes, what values of \(x\) should be included? \(x=\) ________ e. To include the entire block of probability for the first value of \(x=\) ______, start at _______. f. Calculate \(z=\frac{x \pm .5-n p}{\sqrt{n p q}}=\) _______. g. Calculate \(P(x>9) \approx P(z>\)______) \(=1-\) _____ \(=\) ____.

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Find these probabilities for the standard normal random variable \(z\) : a. \(P(z<2.33)\) b. \(P(z<1.645)\) c. \(P(z>1.96)\) d. \(P(-2.58
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