17. It the 1980s, it was generally believed that congenital abnormalities affect
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17. It the 1980s, it was generally believed that congenital abnormalities affected about 77% of a large nation's children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined 380 randomly selected children and found that 43 of them showed signs of an abnormality. Is this strong evidence that the risk has increased? (We consider a P-value of around 55% to represent reasonable evidence.) Complete parts a through f. a) Write appropriate hypotheses. Let p be the proportion of children with genetic abnormalities. Choose the correct answer below. A. H0: pequals=0.070.07 vs. HA: pnot equals0.070.07 B. H0: pequals=0.070.07 vs. HA: pgreater than>0.070.07 C. H0: pequals=0.11320.1132 vs. HA: pnot equals0.11320.1132 D. H0: pequals=0.11320.1132 vs. HA: pless than<0.11320.1132 E. H0: pequals=0.070.07 vs. HA: pless than<0.070.07 F. H0: pequals=0.11320.1132 vs. HA: pgreater than>0.11320.1132 b) Check the necessary assumptions. Which of the following are satisfied? Select all that apply. A. Less than 10% of the population was sampled. B. There are more than 10 successes and 10 failures. C. The sample is random. D. The data are independent. c) Perform the mechanics of the test. What is the P-value? P-valueequals=nothing (Round to three decimal places as needed.) d) Explain carefully what the P-value means in this context. Choose the correct answer below. A. The P-value is the chance of observing 4343 or more children with genetic abnormalities in a random sample of 380380 children. B. The P-value is the chance of observing 4343 or more children with genetic abnormalities in a random sample of 380380 children if 77% of children actually have genetic abnormalities. C. The P-value is the chance of observing 77% of children with genetic abnormalities. D. The P-value is the actual percentage of children who have genetic abnormalities. e) What's your conclusion? A. RejectReject H0. There isis sufficient evidence that more than 77% of the nation's children have genetic abnormalities. B. RejectReject H0. There is notis not sufficient evidence that more than 77% of the nation's children have genetic abnormalities. C. Fail to rejectFail to reject H0. There is notis not sufficient evidence that more than 77% of the nation's children have genetic abnormalities. D. Fail to rejectFail to reject H0. There isis sufficient evidence that more than 77% of the nation's children have genetic abnormalities. f) Do environmental chemicals cause congenital abnormalities? A. No, the conclusion of the hypothesis test shows that environmental chemicals do not cause genetic abnormalities. B. Yes, the conclusion of the hypothesis test shows that environmental chemicals cause genetic abnormalities. C. It is unknown if environmental chemicals cause genetic abnormalities, because the hypothesis test does not indicate the cause of any changes.
18. Data from a recent year showed that 63% of the tens of thousands of applicants to a certain program were accepted. A company that trains applicants claimed that 191 of the 290 students it trained that year were accepted. Assume these trainees were representative of the population of applicants. Has the company demonstrated a real improvement over the average? Complete parts a through c below. a) What are the hypotheses?Which hypotheses below will test if the company demonstrated improvement over the average? A. Upper H 0H0: pequals=0.65860.6586 Upper H Subscript Upper AHA: pless than<0.65860.6586 B. Upper H 0H0: pequals=0.630.63 Upper H Subscript Upper AHA: pless than<0.630.63 C. Upper H 0H0: pequals=0.630.63 Upper H Subscript Upper AHA: pgreater than>0.630.63 D. Upper H 0H0: pequals=0.630.63 Upper H Subscript Upper AHA: pnot equals0.630.63 E. Upper H 0H0: pequals=0.65860.6586 Upper H Subscript Upper AHA: pgreater than>0.65860.6586 F. Upper H 0H0: pequals=0.65860.6586 Upper H Subscript Upper AHA: pnot equals0.65860.6586 b) Verify that the conditions are satisfied and find the P-value. Which assumptions and conditions below are met? Select all that apply. A.The independence assumption is met. B.The success/failure condition is met. C.The 10% condition is met. D.The randomization condition is met. What is the P-value? P-valueequals=nothing (Round to four decimal places as needed.) c) Would you recommend this company based on what you see here? A. The P-value provides strong evidencestrong evidence that the program is successful. B. The P-value provides some indicationsome indication that the program is successful. C. The P-value provides insufficient evidenceinsufficient evidence that the program is successful. D. Inference is not possible since the assumptions and conditions are not met. ____________________________________
19. The manufacturer of a metal stand for home TV sets must be sure that its product will not fail under the weight of the TV. Since some larger sets weigh nearly 300 pounds, the company's safety inspectors have set a standard of ensuring that the stand can support an average of over 500 pounds. Their inspectors regularly subject a random sample of the stands to increasing weight until they fail. They test the hypothesis Upper H 0H0: muequals=500500 against Upper H Subscript Upper AHA: mugreater than>500500, using the level of significance alphaequals=0.01. If the sample of stands fails to pass this safety test, the inspectors will not certify the product for sale to the general public. The manufacturer is thinking of revising its safety test. Complete parts a through c below. a) If the company's lawyers are worried about being sued for selling an unsafe product, should they make the value of alpha smaller or larger? Explain. Choose the correct answer below. A. They should make the value of alpha smaller. This means a smaller chance of declaring the stands unsafe when they are actually safe. B.They should make the value of alpha smaller. This means a smaller chance of declaring the stands safe when they are not actually safe. C.They should make the value of alpha larger. This means a smaller chance of declaring the stands unsafe when they are actually safe. D.They should make the value of alpha larger. This means a smaller chance of declaring the stands safe when they are not actually safe. b) In this context, what is meant by the power of the test? A. The probability of correctly detecting that the stands are not capable of holding more than 500 pounds. B.The probability of incorrectly detecting that the stands are capable of holding more than 500 pounds. C.The probability of correctly detecting that the stands are capable of holding more than 500 pounds. D. The probability of incorrectly detecting that the stands are not capable of holding more than 500 pounds. c) If the company wants to increase the power of the test, what options does it have? Explain the advantages and disadvantages of each option. Select all that apply. A. The company could increase the "design load" to be well above 500500 pounds. This would probably be costly. B. The company could change the production procedure so that the standard deviation of the weights of the stands decreases. This would probably be costly. C. The company could increase the sample size. This would take more time for testing and be costly. D. The company could increase alpha. This would result in more Type I errors.
Explanation / Answer
Solution:-
18)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.659
Alternative hypothesis: P 0.659
Note that these hypotheses constitute a two-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.02785
z = (p - P) /
z = - 1.041
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.041 or greater than 1.041.
Thus, the P-value = 0.2984
Interpret results. Since the P-value (0.2984) is greater than the significance level (0.05), we have to accept the null hypothesis.
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