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For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Testing a new drug with potentially dangerous side effects to see if it is significantly better than the drug currently in use. If it is found to be more effective, it will be prescribed to millions of people.

Short Answer

Expert verified
It is more appropriate to use a relatively small significance level, such as \(\alpha = 0.01\), to minimize the risk of falsely declaring the new drug as more effective.

Step by step solution

01

Understanding the context

In this situation, we are testing a new drug with potentially dangerous side effects to know if it is significantly better than the currently used drug. If we find it more effective, it will be prescribed to millions of people.
02

Interpreting significance level

A significance level (\(\alpha\)) is the probability of rejecting the null hypothesis when it is actually true. Therefore, a larger \(\alpha\) means a greater chance of wrongly rejecting the null hypothesis (Type I error). In the context of this experiment, it would mean a greater chance of concluding that the new drug is more effective when it actually isn't.
03

Conclusion

Considering the stakes, including dangerous side effects and the drug potentially being prescribed to millions of people, a Type I error could have serious consequences. Consequently, it is better to use a relatively small \(\alpha\) (such as \(\alpha = 0.01\)). The lower \(\alpha\) reduces the chance of falsely concluding that the new drug is more effective, therefore minimizing the risk of prescribing an ineffective or potentially harmful drug to millions of people.

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Most popular questions from this chapter

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Using your statistics class as a sample to see if there is evidence of a difference between male and female students in how many hours are spent watching television per week.

In a test to see whether males, on average, have bigger noses than females, the study indicates that " \(p<0.01\)."

Match the four \(\mathrm{p}\) -values with the appropriate conclusion: (a) The evidence against the null hypothesis is significant, but only at the \(10 \%\) level. (b) The evidence against the null and in favor of the alternative is very strong. (c) There is not enough evidence to reject the null hypothesis, even at the \(10 \%\) level. (d) The result is significant at a \(5 \%\) level but not at a \(1 \%\) level. I. 0.00008 II. 0.0571 III. 0.0368 IV. \(\quad 0.1753\)

In Exercise 3.129 on page \(254,\) we see that the home team was victorious in 70 games out of a sample of 120 games in the FA premier league, a football (soccer) league in Great Britain. We wish to investigate the proportion \(p\) of all games won by the home team in this league. (a) Use StatKeyor other technology to find and interpret a \(90 \%\) confidence interval for the proportion of games won by the home team. (b) State the null and alternative hypotheses for a test to see if there is evidence that the proportion is different from 0.5 . (c) Use the confidence interval from part (a) to make a conclusion in the test from part (b). State the confidence level used. (d) Use StatKey or other technology to create a randomization distribution and find the p-value for the test in part (b). (e) Clearly interpret the result of the test using the p-value and using a \(10 \%\) significance level. Does your answer match your answer from part (c)? (f) What information does the confidence interval give that the p-value doesn't? What information does the p-value give that the confidence interval doesn't? (g) What's the main difference between the bootstrap distribution of part (a) and the randomization distribution of part (d)?

Exercises 4.59 to 4.64 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) Sample data: \(\hat{p}=30 / 50=0.60\) with \(n=50\)

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