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Using Technology. In Exercises 5–8, identify the indicated values or interpret the given display. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. Use\(\alpha \)= 0.05 significance level and answer the following:

a. Is the test two-tailed, left-tailed, or right-tailed?

b. What is the test statistic?

c. What is the P-value?

d. What is the null hypothesis, and what do you conclude about it?

e. What is the final conclusion?

Adverse Reactions to Drug The drug Lipitor (atorvastatin) is used to treat high cholesterol. In a clinical trial of Lipitor, 47 of 863 treated subjects experienced headaches (based on data from Pfizer). The accompanying TI@83/84 Plus calculator display shows results from a test of the claim that fewer than 10% of treated subjects experience headaches.

Short Answer

Expert verified

a.The test is left-tailed.

b.The value of the test statistic (z-score) is equal to -4.459.

c.The p-value is equal to 0.00000412.

d. The null hypothesis is that the proportion of treated subjects who experience headaches is equal to 10%.The null hypothesis is rejected.

e. There is enough evidence to conclude that the proportion of treated subjects who experience headaches is less than 10%.

Step by step solution

01

Given information

It is given that out of 863 subjects who were treated with the drug Lipitor, 47 of them experienced headaches. A claim is made that less than 10% of the treated subjects experience headaches.

02

Tail of the test

a.

According to the given claim, the proportion of treated subjects who experience headaches (p) is less than 10%.

Symbolically,\(p < 0.10\).

Asthere is a less than sign in the given claim, the test is left-tailed.

03

Test statistic

b.

The test statistic to test the given claim is the z-score.

Here, the value of the test statistic (z-score) is equal to -4.459.

04

P-value

c.

The p-value corresponding to the z-score of -4.459 is equal to 0.00000412 (in decimals).

05

Null hypothesis and its conclusion

d.

The null hypothesis for this test is as follows:

Null hypothesis: The proportion of treated subjects who experience headaches is equal to 10%.

Symbolically,\({H_0}:p = 0.10\),where

p is the proportion of treated subjects who suffer from headaches.

Here, the p-value equal to 0.00000412 is less than the significance level of 0.05. Thus, the null hypothesis is rejected.

06

Conclusion of the test

e.

There is enough evidence to conclude that the proportion of treated subjects who experience headaches is less than 10%.

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Most popular questions from this chapter

Estimates and Hypothesis Tests Data Set 3 “Body Temperatures” in Appendix B includes sample body temperatures. We could use methods of Chapter 7 for making an estimate, or we could use those values to test the common belief that the mean body temperature is 98.6°F. What is the difference between estimating and hypothesis testing?

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Car Booster Seats The National Highway Traffic Safety Administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement?

774 649 1210 546 431 612

Critical Values. In Exercises 21–24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of α= 0.05, should we reject H0or should we fail to reject H0?

Exercise 19

Finding Critical t Values When finding critical values, we often need significance levels other than those available in Table A-3. Some computer programs approximate critical t values by calculating t=df×eA2/df-1where df = n-1, e = 2.718, A=z8×df+3/8×df+1, and z is the critical z score. Use this approximation to find the critical t score for Exercise 12 “Tornadoes,” using a significance level of 0.05. Compare the results to the critical t score of 1.648 found from technology. Does this approximation appear to work reasonably well?

Technology. In Exercises 9–12, test the given claim by using the display provided from technology. Use a 0.05 significance level. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Old Faithful Data Set 23 “Old Faithful” in Appendix B includes data from 250 random eruptions of the Old Faithful geyser. The National Park Service makes predictions of times to the next eruption, and the data set includes the errors (minutes) in those predictions. The accompanying Statdisk display results from using the prediction errors (minutes) to test the claim that the mean prediction error is equal to zero. Comment on the accuracy of the predictions.

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