6. A company manufacturing computer chips finds that 8% of all chips manufactured are defective. In an effort to decrease the percentage of defective chips, management decides to provide additional training to those employees hired within the last year. After training was implemented, a sample of 450 chips revealed only 27 defects. A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate. Suppose the p-value associated with the test statistic is 0.0594. For a = .01, what is the correct conclusion?

a) The additional training significantly increased the defect rate.

b) The additional training significantly lowered the defect rate.

c) The additional training did affect the defect rate.

d) The additional training did not significantly lower the defect rate.

e) None of these.

7. In the 1980’s, it was generally beloved that congenital abnormalities affected about 7% 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 398 randomly selected children and found that 41 of them showed signs of an abnormality. Is this strong evidence that the risk has increased?

(We consider a P-value of around 5% to represent reasonable evidence).

Complete a through f.

A) Write appropriate hypothesis. Let P be the proportion of children with genetic abnormalities. Choose the correct answer below.

a)

b)

c)

d)

e)

B) Check the necessary assumptions. Which of the following are satisfied?

a) The data are independent.

b) There are more than 10 successes & 10 Failures.

c) Less than 10% of the population was sampled.

d) The sample is random

C) Perform the mechanics of the test. What is the P-value?

P-value = 0.005(Round up to 3 decimal places).

D) Explain carefully what the P-value means in this context. Choose the correct answer.

a) The P-value is the actual percentage of children who have genetic abnormalities.

b) The P-value is the chance of observing 41 or more children with genetic abnormalities in a random sample of 398 children if 7% of children actually have genetic abnormalities.

c) The P-value is the chance of observing 7% of children with genetic abnormalities.

d) The P-value is the chance of observing 41 or more children with genetic abnormalities in a random sample of 398 children.

E) What’s your conclusion?

a) Fail to reject Ho. There is sufficient evidence that more than 7% of the nation’s children have genetic abnormalities.

b) Fail to reject Ho. There is not sufficient evidence that more than 7% of the Nation’s children have genetic abnormalities.

c) Reject Ho. There is not sufficient evidence that more than 7% of the nation’s children have genetic abnormalities.

d) Reject Ho. There is sufficient evidence that more than 7% 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.

8) A survey of 350 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. Suppose that based on the survey, 140 of the 350 students responded “yes”.

a) What is the values of the sample proportion **p^** ?

b) What is the standard error of the sample proportion?

c) Construct an approximate 95% confidence interval for the true proportion p by taking ± 2 SEs from the sample proportion.

a) P^ = 0.4 (type an integer or a decimal)

b) The standard error of the sample proportion is 0.0262

c) The approximate 95% confidence interval for the true proportion p by taking ± 2 SEs from the sample proportion is 0.348,0.452 (round to 3 decimal places)

9) A biotechnology firm is planning its investment strategy for future products and research labs. A poll found that 12% of a random sample of 1034 adults approved of attempts to clone a human. Use this information to complete parts a through E.

A) Find the margin of error for this poll if we want 99% confidence in our estimate of the percent of adults who approves of cloning humans.

ME = 0.026(ROUND TO 3 DECIMAL PLACES)

B) Explain what the margin of error means:

a) The pollsters are 99% confident that the margin of error contains the true proportion of adults who approve of attempts to clone a human.

b) The margin of error is the width of the confidence interval that contains the true proportion of adults who approve of attempts to clone a human

c) The pollsters are 99% confident that the true proportion of adults who approve of attempts to clone a human is within the margin of error of the estimated 12%

d) The mean of error is the value that should be subtracted from the 99% confidence level to obtain the pollsters true confidence level.

C) If we only need to be 95% confident, will the margin of error be larger or smaller?

a) A 95% confidence interval requires a larger margin and error. In order to increase confidence, the interval must be wider.

b) A 95% confidence interval requires a smaller margin of error. A narrower interval leads to decrease confidence.

D) Find that margin of error.

ME = 0.20

E) In general, if all other aspects of the situation remain the same, would smaller samples produce small or larger margin of error?

a) Smaller samples produce smaller margins of error.

b) Smaller samples produce larger margins of error.

10) For parts a & b, use the **t** tables, software, or calculator to estimate.

**A) **The critical value of **t **for a 99% confidence interval with **df**=20 is 2.85 Round to 2 decimal places.

B) The critical value of **t** for a 90% confidence interval with **df**=62 is 1.67 Round to two decimal places.

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