﻿ ﻿Inference for Means and Proportions - Review the Knowledge You Need to Score High - 5 Steps to a 5 AP Statistics 2017 (2016)

## 5 Steps to a 5 AP Statistics 2017 (2016)

### Inference for Means and Proportions

IN THIS CHAPTER

Summary: In the last chapter, we concentrated on estimating, using a confidence interval, the value of a population parameter or the difference between two population parameters. In this chapter, we test to see whether some specific hypothesized value of a population parameter, or the difference between two populations parameters, is likely or not. We form hypotheses and test to determine the probability of getting a particular finding if the null hypothesis is true. We then make decisions about our hypothesis based on that probability.

Key Ideas

The Logic of Hypothesis Testing

z -Procedures versus t -Procedures

Inference for a Population Mean

Inference for the Difference between Two Population Means

Inference for a Population Proportion

Inference for the Difference between Two Population Proportions

We began our study of inference in the previous chapter by using confidence intervals to estimate population values. We estimated a single population mean and a single population proportion as well as the difference between two population means and between two population proportions. We discussed z- and t -procedures for estimating a population mean. In this chapter, we build on those techniques to evaluate claims about population values.

Significance Testing

Before we actually work our way through inference for means and proportions, we need to review the hypothesis testing procedure introduced in Chapter 11 and to understand how questions involving inference will be scored on the AP exam. In the previous chapter, we identified the steps in the hypothesis-testing procedure as follows:

1. State hypotheses in the context of the problem. The first hypothesis, thenull hypothesis , is the hypothesis we are actually testing. The null hypothesis usually states that there is nothing going on: the claim is correct or there is no distinction between groups. It is symbolized by H 0 .

The second hypothesis, the alternative hypothesis , is the theory that the researcher wants to confirm by rejecting the null hypothesis. The alternative hypothesis is symbolized by H A . There are three forms for the alternative hypothesis: ≠, >, or <. That is, if the null is H 0 : μ 1μ 2 = 0, then H A could be:

(In the case of the one-sided alternative H A : μ 1μ 2 > 0, the null hypothesis is sometimes written H 0 : μ 1μ 2 ≤ 0.)

1. Identify which procedure you intend to use and show that the conditions for its use are satisfied. If you are going to state a significance level, α, it can be done here.

III. Compute the value of the test statistic and the P -value.

1. Using the value of the test statistic and/or theP -value, give a conclusion in the context of the problem.

If you stated a significance level, the conclusion can be based on a comparison of the P -value with α. If you didn”t state a significance level, you can argue your conclusion based on the relative size of the P -value alone: if it is small, you have evidence against the null hypothesis; if it is not small, you do not have evidence against the null hypothesis.

The conclusion can be (1) that we reject H 0 (because of a sufficiently small P -value) or (2) that we do not reject H 0 (because the P -value is too large). We never accept the null hypothesis: we either reject it or fail to reject it. If we reject H 0 , we can say that we have evidence in favor of H A .

Significance testing involves making a decision about whether or not an observed result is statistically significant. That is, is the result sufficiently unlikely, if the null hypothesis were true, so as to provide good evidence for rejecting the null hypothesis in favor of the alternative? The four steps in the hypothesis testing process outlined above are the four steps that are required on the AP exam when doing inference problems. In brief, every test of a hypothesis should have the following four steps:

1. State the null and alternative hypotheses in the context of the problem, defining all symbols.
2. Identify the appropriate test and check that the conditions for its use are present.

III. Do the correct mechanics, including calculating the value of the test statistic and the P -value.

1. State a correct conclusion in the context of the problem.

Exam Tip: You are not required to number the four steps on the exam, but it is a good idea to do so—then you are sure you have done everything required. Note that the correct mechanics are only worth about 25% of the problem.

z -Procedures versus t -Procedures

In this chapter, we explore inference for means and proportions. When we deal with means, we use t -procedures, unless there is some unusual situation in which s is known. If that were to happen, use z -procedures. With proportions, assuming the proper conditions are met, we deal only with large samples—that is, with z -procedures.

When doing inference for a population mean, or for the difference between two population means, we will usually use t -procedures. This is because z -procedures assume that we know the population standard deviation (or deviations in the two-sample situation), which we rarely do. We typically use t -procedures when doing inference for a population mean or for the difference between two population means when:

(a) The sample is a simple random sample from the population

and

(b) The sample size is large (rule of thumb: n ≥ 30) or the population from which the sample is drawn is approximately normally distributed (or, at least, does not depart dramatically from normal).

You could use z -procedures when doing inference for population means if:

(a) The samples are simple random samples from the population

and

(b) The population(s) from which the sample(s) is (are) drawn is normally distributed (in this case, the sampling distribution of or x 1x 2 will also be normally distributed)

and

(c) The population standard deviation(s) is (are) known. This would be a rare situation and is unlikely to be seen on the AP Exam.

Historically, many texts allowed you to use z procedures when doing inference for means if your sample size was large enough to argue, based on the central limit theorem, that the sampling distribution of or 12 is approximately normal. The basic assumption is that, for large samples, the sample standard deviation s is a reasonable estimate of the population standard deviation σ . Today, most statisticians would tell you that it”s better practice to always use t -procedures when doing inference for a population mean or for the difference between two population means. You can receive credit on the AP exam for doing a large sample problem for means using z -procedures if you specify that you are doing so because the sample size is large, but it”s definitely better practice to use t -procedures.

When using t -procedures, it is important to check in step II of the hypothesis test procedure that the data could plausibly have come from an approximately normal population. A stemplot, boxplot, dotplot, or normal probability plot can be used to show there are no outliers or extreme skewness in the data. t -procedures are robust against these assumptions, which means that the procedures still work reasonably well even with some violation of the condition of normality, provided there is not much skewness. Some texts use the following guidelines for sample size when deciding whether of not to use t -procedures:

• n< 15. Use t -procedures if the data show no outliers and no skewness.
• 15 <n < 40. Use t -procedures unless there are outliers or marked skewness.
• n> 40. Use t -procedures for any distribution.

For the two-sample case discussed later, these guidelines can still be used if you replace n with n 1 and n 2 .

Using Confidence Intervals for Two-Sided Alternatives

Consider a two-sided significance test at, say, α = 0.05 and a confidence interval with C = 0.95. A sample statistic that would result in a significant result at the 0.05 level would also generate a 95% confidence interval that does not contain the hypothesized value. Confidence intervals for two-sided hypothesis tests could then be used in place of generating a test statistic and finding a P -value. If the sample value generates a C -level confidence interval that does not contain the hypothesized value of the parameter, then a significance test based on the same sample value would reject the null hypothesis at α = 1 – C . Questions sometimes ask for such a decision to be made based on a confidence interval. Beware, however. If a question simply asks if there is statistical evidence to support a hypothesis it is asking for a test. It is best to do what is asked for.

You should never use confidence intervals for hypothesis tests involving one-sided alternative hypotheses. For the purposes of this book, confidence intervals are considered to be two sided. (One-sided confidence intervals are not a part of this course.)

Inference for a Single Population Mean

In step II of the hypothesis-testing procedure, we need to identify the test to be used and justify the conditions needed. The test can be identified by name or by formula. For example, you could say “We will do a significance test for a mean” to identify the procedure in words. To identify the procedure with a formula, write the formula for the test statistic which will usually have the following form:

When doing inference for a single mean, the estimator is , the hypothesized value is μ 0 in the null hypothesis H 0 : μ = μ 0 , and the standard error is the estimate of the standard deviation of , which is

This can be summarized in the following table.

example: A study was done to determine if 12- to 15-year-old girls who want to be engineers differ in IQ from the average of all girls. The mean IQ of all girls in this age range is known to be about 100 with a standard deviation of 15. A random sample of 49 girls who state that they want to be engineers is selected and their IQ is measured. The mean IQ of the girls in the sample is 104.5. Does this finding provide evidence, at the 0.05 level of significance, that the mean IQ of 12- to 15-year-old girls who want to be engineers differs from the average? Assume that the population standard deviation is 15 (σ = 15).

solution 1 (test statistic approach): The solution to this problem will be put into a form that emphasizes the format required when writing out solutions on the AP exam.

I . Let μ = the true mean IQ of girls who want to be engineers.

H 0 : μ = 100.

H A : μ ≠ 100.

(The alternative is two-sided because the problem wants to know if the mean IQ “differs” from 100. It would have been one-sided if it had asked whether the mean IQ of girls who want to be engineers is higher than average.)

II . Since σ is known, we will use a one-sample z -test at α = 0.05.

Conditions:

• The problem states that we have a random sample.
• Sample size is large.
• sis known.

III . ⇒ P -value = 2(1–0.9821) = 0.0358 (from Table A).

The TI-83/84 gives 2 × normalcdf(2.10,100)=0.0357.

IV . Because P < α, we reject H 0 . We have strong evidence that the true mean IQ for girls who want to be engineers differs from the mean IQ of all girls in this age range.

Notes on the above solution:

• Had the alternative hypothesis been one-sided, theP -value would have been 1 – 0.9821 = 0.0179. We multiplied by 2 in step III because we needed to consider the area in both tails.
• The problem told us that the significance level was 0.05. Had it not mentioned a significance level, we could have arbitrarily chosen one, or we could have argued a conclusion based only on the derivedP -value without a significance level.
• The linkage between theP -value and the significance level must be made explicit in part IV. Some sort of statement, such as “Since P = 0.0358 < α = 0.05 …” or, if no significance level is stated, “Since the P -value is low …” will indicate that your conclusion is based on the P -value determined in step III.

Calculator Tip: The TI-83/84 can do each of the significance tests described in this chapter as well as those in Chapters 13 and 14 . It will also do all of the confidence intervals we considered in Chapter 11 . You get to the tests by entering STAT TESTS . Note that Z-Test and T-Test are what are often referred to as “one-sample tests.” The two-sample tests are identified as 2-SampZTest and 2-SampTTest . Most of the tests will give you a choice between Dataand Stats . Data means that you have your data in a list and Stats means that you have the summary statistics of your data. Once you find the test you are interested in, just follow the calculator prompts and enter the data requested.

solution 2 (confidence interval approach—ok since H A is two-sided, but not recommended):

1. Let μ = the true mean IQ of girls who want to be engineers.

H 0 : μ = 100.

H A : μ ≠ 100.

1. We will use a 95%z -confidence interval (C = 0.95).

Conditions:

• We have a large sample size (n= 49).
• The standard deviation is known (σ= 15).

III. = 104.5, z * = 1.96.

We are 95% confident that the true population mean is in the interval (100.3, 108.7).

1. Because 100 is not in the 95% confidence interval forµ , we reject H 0 . We have strong evidence that the true mean IQ for girls who want to be engineers differs from the mean IQ of all girls in this age range.

example: A company president believes that there are more absences on Monday than on other days of the week. The company has 45 workers. The following table gives the number of worker absences on Mondays and Wednesdays for a random sample of 8-weeks from the past two years. Do the data provide evidence that there are more absences on Mondays?

solution: Because the data are paired on a weekly basis, the data we use for this problem are the difference between the days of the week for each of the 8 weeks. Adding a row to the table that gives the differences (absences on Monday minus absences on Wednesday), we have:

I . Let μ d = the true mean difference between number of absences on Monday and absences on Wednesday.

H 0 : μ d = 0(or H 0 : μ d ≤ 0).

H A : μ d > 0.

II . We will use a one-sample t -test for the difference scores.

Conditions:

A boxplot of the data shows no significant departures from normality (no outliers or severe skewness). Also, we are told that the 8-weeks are randomly selected.

III . = 1.75, s = 2.49, , df = 8 – 1 = 7.

0.025 < P < 0.05 (from Table A).

Using the T-Test function in the STAT TESTS menu of the TI-83/84, the P -value = 0.044.

IV . The P -value is small. This provides us with evidence that, over the past two years, there have been more absences on Mondays than on Wednesdays.

Inference for the Difference Between Two Population Means

The two-sample case for the difference in the means of two independent samples is more complicated than the one-sample case. The hypothesis testing logic is identical, however, so the differences are in the mechanics needed to do the problems, not in the process. For hypotheses about the differences between two means, the procedures are summarized in the following table.

• The second rows for conditions and test statistic give the typicalt -procedure approach. The “software” method is based on the formula given in Chapter 11 . The second way of computing degrees of freedom is the “conservative” method introduced in Chapter 11 .

example: A statistics teacher, Mr. Srednih, gave a quiz to his 8:00 AM class and to his 9:00 AM class. There were 50 points possible on the quiz. The data for the two classes were as follows.

Before the quiz, some members of the 9:00 AM class had been bragging that later classes do better in statistics. Considering these two classes as random samples from the populations of 8:00 AM and 9:00 AM classes, do these data provide evidence at the .01 level of significance that students in 9:00 AM classes do better than those in 8:00 AM classes?

solution:

I . Let μ 1 = the true mean score for the 8:00 AM class.

Let μ 2 = the true mean score for the 9:00 AM class.

H 0 : μ 1μ 2 = 0.

H A : μ 1μ 2 < 0.

II . We will use a two-sample t -test at α = 0.01. We assume these samples are random samples from independent populations. t -procedures are justified because both sample sizes are larger than 30.

(Note: If the sample sizes were not large enough, we would need to know that the samples were drawn from populations that are approximately normally distributed.)

III. , df = min{34 – 1, 31 – 1} = 30.

From Table B, 0.005 < P -value < 0.01. Using the tcdf function from the DISTR menu of the TI-83/84 yields tcdf(-100,-2.53,30)=0.0084 . Using the 2SampTTest function in the STAT TESTS menu of the TI-83/84 for the whole test yields t = –2.53 and a P -value of 0.007 based on df = 49.92 (the lower P -value being based on a larger number of degrees of freedom).

IV . Because P < 0.01, we reject the null hypothesis. We have good evidence that the true mean for 9:00 AM classes is higher than the true mean for 8:00 AM classes.

Inference for a Single Population Proportion

We are usually more interested in estimating a population proportion with a confidence interval than we are in testing that a population proportion has a particular value. However, significance testing techniques for a particular population proportion exist and follow a pattern similar to those of the previous two sections. The main difference is that the only test statistic is z . The logic is based on using a normal approximation to the binomial as discussed in Chapter 10 .

Notes on the preceding table:

• The standard error for a hypothesis test of a single population proportion is different from that for a confidence interval for a population proportion. The standard error for a confidence interval,

is a function of , the sample proportion, whereas the standard error for a hypothesis test,

is a function of the hypothesized value of p . This is because a test assumes, for the sake of our model, that p = p 0 . So we use that in the calculation of the standard error. An interval makes no such assumption.

• Like the conditions for the use of az -interval, we require that the np 0 and n (1 –p 0 ) be large enough to justify the normal approximation. As with determining the standard error, we use the hypothesized value of p rather than the sample value. “Large enough” means either np 0 ≥ 5 and n (1 –p 0 ) ≥ 5, or np 0 ≥ 10 and n (1 –p 0 ) ≥ 10 (it varies by text).

example: Consider a screening test for prostate cancer that its maker claims will detect the cancer in 85% of the men who actually have the disease. One hundred seventy-five men who have been previously diagnosed with prostate cancer are given the screening test, and 141 of the men are identified as having the disease. Does this finding provide evidence that the screening test detects the cancer at a rate different from the 85% rate claimed by its manufacturer?

solution:

I . Let p = the true proportion of men with prostate cancer who test positive.

H 0 : p = 0.85.

H A : p ≠ 0.85.

II . We want to use a one-proportion z -test. 175(0.85) = 148.75 > 5 and 175(1 – 0.85) = 26.25 > 5, so the conditions are present to use this test (the conditions are met whether we use 5 or 10). There is no randomization here. But we can do the test to see if the data provide evidence that, among these patients, the difference from 0.85 is too big to be reasonably attributed to random variation.

III .

Using the TI-83/84, the same answer is obtained by using 1-PropZTest in the STAT TESTS menu.

IV . Because P is reasonably large, we do not have sufficient evidence to reject the null hypothesis. The evidence is insufficient to challenge the company”s claim that the test is 85% effective.

example: Maria has a quarter that she suspects is out of balance. In fact, she thinks it turns up heads more often than it would if it were fair. Being a senior, she has lots of time on her hands, so she decides to flip the coin 300 times and count the number of heads. There are 165 heads in the 300 flips. Does this provide evidence at the 0.05 level that the coin is biased in favor of heads? At the 0.01 level?

solution:

I . Let p = the true proportion of heads in 300 flips of a fair coin.

H 0 : p = 0.50 (or H 0 : p ≤ 0.50).

H A : p > 0.50.

II . We will use a one-proportion z -test.

300(0.50) = 150 > 5 and 300(1 – 0.50) = 150 > 5, so the conditions are present for a one-proportion z -test.

III . .

(This can also be done using 1-PropZTest in the STAT TESTS menu of the TI-83/84.)

IV . The question asked about both α = 0.05 and α = 0.01. We note that this finding would have been significant at α = 0.05 but not at α = 0.01. For the two significance levels, the solutions would look something like this:

(1) Since the P -value = 0.042 < α = 0.05, we reject the null hypothesis. We have evidence that the true proportion of heads is greater than 0.50.

(2) Since the P -value = 0.042 > α = 0.01, we fail to reject the null hypothesis. We do not have convincing evidence that the true proportion of heads is greater than 0.50.

Inference for the Difference Between Two Population Proportions

The logic behind inference for two population proportions is the same as for the others we have studied. As with the one-sample case, there are some differences between z -intervals and z -tests in terms of the computation of the standard error. The following table gives the essential details.

example: Two concentrations of a new vaccine designed to prevent infection are developed and a study is conducted to determine if they differ in their effectiveness. The results from the study are given in the following table.

Does this study provide statistical evidence at the α = 0.01 level that the vaccines differ in their effectiveness?

solution:

I . Let p 1 = the population proportion infected after receiving Vaccine A.

Let p 2 = the population proportion infected after receiving Vaccine B.

H 0 : p 1p 2 = 0.

H A : p 1p 2 ≠ 0.

II . We will use a two-proportion z -test at α = 0.01.

n 1 1 = 225(0.453), = 102, n 1 (1 – 1 ) = 225(0.547) = 123,

n 2 2 = 285(0.333) = 95 n 2 = 2 = 225(0.667) = 190.

All values are larger than 5, so the conditions necessary are present for the two-proportion z -test.

(Note: When the counts are given in table form, as in this exercise, the values found in the calculation above are simply the table entries! Take a look. Being aware of this could save you some work.)

III .

0.006 (from Table A). (Remember that you have to multiply by 2 since it is a two-sided alternative—you are actually finding the probability of being 2.76 standard deviations away from the mean in some direction.) Given the z -score, the P -value could also be found using a TI-83/84: 2 × normalcdf (2.76,100).

(Using a TI-83/84, all the mechanics of the exercise could have been done using the 2-PropZTest in the STAT TESTS menu. If you do that, remember to show enough work that someone reading your solution can see where the numbers came from.)

IV . Because P < 0.01, we have grounds to reject the null hypothesis. We have strong evidence that the two vaccines differ in their effectiveness. Although this was two-sided test, we note that Vaccine A was less effective than Vaccine B.

Rapid Review

1. A researcher reports that a finding of = 3.1 is significant at the 0.05 level of significance. What is the meaning of this statement?

Answer: Under the assumption that the null hypothesis is true, the probability of getting a value at least as extreme as the one obtained is less than 0.05. It was unlikely to have occurred by chance.

1. Letμ 1 = the mean score on a test of agility using a new training method and let μ 2 = the mean score on the test using the traditional method. Consider a test of H 0 : μ 1μ 2 = 0. A large sample significance test finds P = 0.04. What conclusion, in the context of the problem, do you report if

(a) α = 0.05?

(b) α = 0.01?

(a) Because the P -value is less than 0.05, we reject the null hypothesis. We have evidence that there is a non-zero difference between the traditional and new training methods.

(b) Because the P -value is greater than 0.01, we do not have sufficient evidence to reject the null hypothesis. We do not have strong support for the hypothesis that the training methods differ in effectiveness.

1. True–False: In a hypothesis test concerning a single mean, we can use eitherz- procedures or t- procedures as long as the sample size is at least 20.

Answer: False. With a sample size of only 20, we can not use z -procedures unless we know that the population from which the sample was drawn is approximately normal and σ is known. We can use t -procedures if the data do not have outliers or severe skewness, that is, if the population from which the sample was drawn is approximately normal.

1. We are going to conduct a two-sided significance test for a population proportion. The null hypothesis isH 0 : p = 0.3. The simple random sample of 225 subjects yields = 0.35. What is the standard error, s , involved in this procedure if

(a) you are constructing a confidence interval for the true population proportion?

(b) you are doing a significance test for the null hypothesis?

(a) For a confidence interval, you use the value of in the standard error.

Hence,

(b) For a significance test, you use the hypothesized value of p . Hence,

1. For the following data,

(a) justify the use of a two-proportion z -test for H 0 : p 1p 2 = 0.

(b) what is the value of the test statistic for H 0 : p 1p 2 = 0?

(c) what is the P -value of the test statistic for the two-sided alternative?

(a) n 1 1 = 40(0.3) = 12, n 1 (1 – 1 ) = 40(0.7) = 28,

n 2 2 = 35(0.4) = 14, n 2 (1 – 2 ) = 35(0.6) = 21.

Since all values are at least 5, the conditions are present for a two-proportion z -test.

(b)

.

(c) z = –0.91, P -value = 2(0.18) = 0.36 (from Table A). On the TI-83/84, this P -value can be found as 2 × normalcdf(-100,-0.91) .

1. You want to conduct a one-sample test (t-test) for a population mean. Your random sample of size 10 yields the following data: 26, 27, 34, 29, 38, 30, 28, 30, 30, 23. Should you proceed with your test? Explain.

Answer: A boxplot of the data shows that the 38 is an outlier. Further, the dotplot of the data casts doubt on the approximate normality of the population from which this sample was drawn. Hence, you should not use a t -test on these data.

1. Although it may be difficult to justify, there are conditions under which you canpool your estimate of the population standard deviation when doing a two-sample test for the difference between population means. When is this procedure justified? Why is it difficult to justify?

Answer: This procedure is justified when you can assume that the population variances (or population standard deviations) are equal. This is hard to justify because of the lack of a strong statistical test for the equality of population variances.

Practice Problems

Multiple-Choice

1. A school district claims that the average teacher in the district earns \$48,000 per year. The teachers” organization argues that the average salary is less. A random sample of 25 teachers yields a mean salary of \$47,500 with a sample standard deviation of \$2000. Assuming that the distribution of all teachers” salaries is approximately normally distributed, what is the value of the t -test statistic and the P -value for a test of the hypothesis H 0 : μ = 48,000 against H A : μ < 48,000?
2. t= 1.25, 0.10 < P < 0.15
3. t= –1.25, 0.20 < P < 0.30
4. t= 1.25, 0.20 < P < 0.30
5. t= –1.25, 0.10 < P < 0.15
6. t= –1.25, P > 0.25
7. Which of the following conditions is (are) necessary to justify the use of z -procedures in a significance test about a population proportion?
8. The samples must be drawn from a normal population.
9. The population must be much larger (10–20 times) than the sample.

III. np 0 ≥ 5 and n (1 – p 0 ) ≥ 5.

1. I only
2. I and II only
3. II and III only
4. III only
5. I, II, and III
6. A minister claims that more than 70% of the adult population attends a religious service at least once a month. Let p = the proportion of adults who attend church. The null and alternative hypotheses you would use to test this claim would be:
7. H0 : p ≤ 0.7, H A : p > 0.7
8. H0 : μ ≤ 0.7, H A : μ > 0.7
9. H0 : p = 0.7, H A : p ≠ 0.7
10. H0 : p ≤ 0.7, H A : p < 0.7
11. H0 : p ≥ 0.7, H A : p < 0.7
12. A t -test for the difference between two populations, means is to be conducted. The samples, of sizes 12 and 15, are considered to be random samples from independent, approximately normally distributed, populations. Which of the following statements is (are) true?
13. If we can assume the population variances are equal, the number of degrees of freedom is 25.
14. An appropriate conservative estimate of the number of degrees of freedom is 11.

III. The P -value for the test statistic in this situation will be larger for 11 degrees of freedom than for 25 degrees of freedom.

1. I only
2. II only
3. III only
4. I and II only
5. I, II, and III
6. When is it OK to use a confidence interval instead of computing a P -value in a hypothesis test?
7. In any significance test
8. In any hypothesis test with a two-sided alternative hypothesis
9. Only when the hypothesized value of the parameter isnot in the confidence interval
10. Only when you are conducting a hypothesis test with a one-sided alternative
11. Only when doing a test for a single population mean or a single population proportion
12. Which of the following is not a required step for a significance test?
13. State null and alternative hypotheses in the context of the problem.
14. Identify the test to be used and justify the conditions for using it.
15. State the significance level for which you will decide to reject the null hypothesis.
16. Compute the value of the test statistic and theP- value.
17. State a correct conclusion in the context of the problem.
18. Which of the following best describes what we mean when say that t -procedures are robust?
19. Thet -procedures work well with almost any distribution.
20. The numerical value oft is not affected by outliers.
21. Thet -procedures will still work reasonably well even if the assumption of normality is violated.
22. t-procedures can be used as long as the sample size is at least 40.
23. t-procedures are as accurate as z -procedures.
24. For a hypothesis test of H 0 : μ = μ 0 against the alternative H A : μ < μ 0 , the z -test statistic is found to be 2.00. This finding is
25. significant at the 0.05 level but not at the 0.0l level.
26. significant at the 0.01 level but not at the 0.05 level.
27. significant at both the 0.01 and the 0.05 levels.
28. significant at neither the 0.01 nor the 0.05 levels.
29. not large enough to be considered significant.
30. Two types of tennis balls were tested to determine which one goes faster on a serve. Eight different players served one of each type of ball and the results were recorded:

Assuming that the speeds are approximately normally distributed, how many degrees of freedom will there be in the appropriate t -test used to determine which type of tennis ball travels faster?

1. 6
2. 7
3. 16
4. 15
5. 14
6. Two statistics teachers want to compare their teaching methods. They decide to give the same final exam and use the scores on the exam as a basis for comparison. They decide that the value of interest to them will be the proportion of students in each class who score above 80% on the final. One class has 32 students and one has 27 students. Which of the following would be the most appropriate test for this situation?
7. Two proportionz -test
8. Two-samplet -test
9. Chi-square goodness-of-fit test
10. One-samplez -test
11. Chi-square test for independence

Free-Response

1. A large high school has been waging a campaign against drug use, particularly marijuana. Before the campaign began in 2004, a random sample of 100 students from the junior and senior classes found 27 who admitted to using marijuana (we understand that some students who used marijuana would be reluctant to admit it on a survey). To assess the success of their program, in early 2007 they surveyed a random sample of 175 juniors and seniors and 30 responded that they have used marijuana. Is this good evidence that the use of marijuana has been reduced at the school?
2. Twenty-six pairs of identical twins are enrolled in a study to determine the impact of training on ability to memorize a string of letters. Two programs (A and B) are being studied. One member of each pair is randomly assigned to one of the two groups and the other twin goes into the other group. Each group undergoes the appropriate training program, and then the scores for pairs of twins are compared. The means and standard deviations for groups A and B are determined as well as the mean and standard deviation for the difference between each twin”s score. Is this study a one-sample or two-sample situation, and how many degrees of freedom are involved in determining the t -value?
3. Which of the following statements is (are) correct? Explain.
4. A confidence interval can be used instead of a test statistic in any hypothesis test involving means or proportions.
5. A confidence interval can be used instead of a test statistic in a two-sided hypothesis test involving means or proportions.

III. The standard error for constructing a confidence interval for a population proportion and the standard error for a significance test for a population proportion are the same.

1. The standard error for constructing a confidence interval for a population mean and the standard error for a significance test for a population mean are the same.
2. The average math SAT score at Hormone High School over the years is 520. The mathematics faculty believes that this year”s class of seniors is the best the school has ever had in mathematics. One hundred seventy-five seniors take the exam and achieve an average score of 531 with a sample standard deviation of 96. Does this performance provide good statistical evidence that this year”s class is, in fact, superior?
3. An avid reader, Booker Worm, claims that he reads books that average more than 375 pages in length. A random sample of 13 books on his shelf had the following number of pages: 595, 353, 434, 382, 420, 225, 408, 422, 315, 502, 503, 384, 420. Do the data support Booker”s claim? Test at the 0.05 level of significance.
4. The statistics teacher, Dr. Tukey, gave a 50-point quiz to his class of 10 students and they didn”t do very well, at least by Dr. Tukey”s standards (which are quite high). Rather than continuing to the next chapter, he spent some time reviewing the material and then gave another quiz. The quizzes were comparable in length and difficulty. The results of the two quizzes were as follows.

Do the data indicate that the review was successful, at the .05 level, at improving the performance of the students on this material? Give statistical evidence for your conclusion.

1. The new reality TV show, “I Want to Marry a Statistician,” has been showing on Monday evenings, and ratings show that it has been watched by 55% of the viewing audience each week. The producers are moving the show to Wednesday night but are concerned that such a move might reduce the percentage of the viewing public watching the show. After the show has been moved, a random sample of 500 people who are watching television on Wednesday night are surveyed and asked what show they are watching. Two hundred fifty-five respond that they are watching “I Want to Marry a Statistician.” Does this finding provide evidence at the 0.01 level of significance that the percentage of the viewing public watching “I Want to Marry a Statistician” has declined?
2. Harvey is running for student body president. A survey is conducted by the AP Statistics class in an attempt to predict the outcome of the election. They randomly sample 30 students, 16 of whom say they plan to vote for Harvey. Harvey figures (correctly) that 53.3% of students in the sample intend to vote for him and is overjoyed at his soon-to-be-celebrated victory. Explain carefully why Harvey should not get too excited until the votes are counted.
3. A company uses two different models, call them model A and model B, of a machine to produce electronic locks for hotels. The company has several hundred of each machine in use in its various factories. The machines are not perfect, and the company would like to phase out of service the one that produces the most defects in the locks. A random sample of 13 model A machines and 11 model B machines are tested and the data for the average number of defects per week are given in the following table.

Dotplots of the data indicate that there are no outliers or strong skewness in the data and that there are no strong departures from normal. Do these data provide statistically convincing evidence that the two machines differ in terms of the number of defects produced?

1. Take another look at the preceding problem. Suppose there were 30 of each model machine that were sampled. Assuming that the sample means and standard deviations are the same as given in question 9, how might this have affected the hypothesis test you performed in that question?
2. The directors of a large metropolitan airport claim that security procedures are 98% accurate in detecting banned metal objects that passengers may try to carry onto a plane. The local agency charged with enforcing security thinks the security procedures are not as good as claimed. A study of 250 passengers showed that screeners missed nine banned carry-on items. What is the P -value for this test and what conclusion would you draw based on it?
3. A group of 175 married couples are enrolled in a study to see if women have a stronger reaction than men to videos that contain violent material. At the conclusion of the study, each couple is given a questionnaire designed to measure the intensity of their reaction. Higher values indicate a stronger reaction. The means and standard deviations for all men, all women, and the differences between husbands and wives are as follows:

Do the data give strong statistical evidence that wives have a stronger reaction to violence in videos than do their husbands? Assume that σ for the differences is 1.77.

1. An election is bitterly contested between two rivals. In a poll of 750 potential voters taken 4 weeks before the election, 420 indicated a preference for candidate Grumpy over candidate Dopey. Two weeks later, a new poll of 900 randomly selected potential voters found 465 who plan to vote for Grumpy. Dopey immediately began advertising that support for Grumpy was slipping dramatically and that he was going to win the election. Statistically speaking (say at the 0.05 level), how happy should Dopey be (i.e., how sure is he that support for Grumpy has dropped)?
2. Consider, once again, the situation of question #7 above. In that problem, a one-sided, two-proportion z -test was conducted to determine if there had been a drop in the proportion of people who watch the show “I Want to Marry a Statistician” when it was moved from Monday to Wednesday evenings.

Suppose instead that the producers were interested in whether the popularity ratings for the show had changed in any direction since the move. Over the seasons the show had been broadcast on Mondays, the popularity rating for the show (10 high, 1 low) had averaged 7.3. After moving the show to the new time, ratings were taken for 12 consecutive weeks. The average rating was determined to be 6.1 with a sample standard deviation of 2.7. Does this provide evidence, at the 0.05 level of significance, that the ratings for the show have changed? Use a confidence interval, rather than a t -test, as part of your argument. A dotplot of the data indicates that the ratings are approximately normally distributed.

1. A two-sample study for the difference between two population means will utilize t -procedures and is to be done at the 0.05 level of significance. The sample sizes are 23 and 27. What is the upper critical value (t *) for the rejection region if

(a) the alternative hypothesis is one-sided, and the conservative method is used to determine the degrees of freedom?

(b) the alternative hypothesis is two-sided and the conservative method is used to determine the degrees of freedom?

(c) the alternative hypothesis is one-sided and you assume that the population variances are equal?

(d) the alternative hypothesis is two-sided, and you assume that the population variances are equal?

Cumulative Review Problems

1. How large a sample is needed to estimate a population proportion within 2.5% at the 99% level of confidence if
2. you have no reasonable estimate of the population proportion?
3. you have data that show the population proportion should be about 0.7?
4. Let X be a binomial random variable with n = 250 and p = 0.6. Use a normal approximation to the binomial to approximate P (X > 160).
5. Write the mathematical expression you would use to evaluate P (X > 2) for a binomial random variable X that has B (5, 0.3) (that is, X is a binomial random variable equal to the number of successes out of 5 trials of an event that occurs with probability of success p = 0.3). Do not evaluate.
6. An individual is picked at random from a group of 55 office workers. Thirty of the workers are female, and 25 are male. Six of the women are administrators. Given that the individual picked is female, what is the probability she is an administrator?
7. A random sample of 25 cigarettes of a certain brand were tested for nicotine content, and the mean was found to be 1.85 mg with a standard deviation of 0.75 mg. Find a 90% confidence interval for the mean number of mg of nicotine in this type of cigarette. Assume that the amount of nicotine in cigarettes is approximately normally distributed. Interpret the interval in the context of the problem.

Solutions to Practice Problems

Multiple-Choice

1. The correct answer is (d).

for the one-sided alternative. The calculator answer is P = 0.112. Had the alternative been two-sided, the correct answer would have been (b).

1. The correct answer is (c). If the sample size conditions are met, it is not necessary that the samples be drawn from a normal population.
2. The correct answer is (a). Often you will see the null written as H 0 : p = 0.7 rather than H 0 : p ≤ 0.7. Either is correct.
3. The correct answer is (e). If we can assume that the variances are equal, then df = n 1 + n 2 – 2 = 12 + 15 – 2 = 25. A conservative estimate for the number of degrees of freedom is df = min {n 1 – 1, n 2 – 1} = min {12 – 1, 25 – 1} = 11. For a given test statistic, the greater the number of degrees of freedom, the lower the P -value.
4. The correct answer is (b). In AP Statistics, we consider confidence intervals to be two-sided. A two-sided α-level significance test will reject the null hypothesis whenever the hypothesized value of the parameter is not contained in the C = 1 – α level confidence interval.

6 (c) is not one of the required steps. You can state a significance level that you will later compare the P- value with, but it is not required. You can simply argue the strength of the evidence against the null hypothesis based on the P- value alone—small values of P provide evidence against the null.

1. (c) is the most correct response. (a) is incorrect because t- procedures do not work well with, for example, small samples that come from non-normal populations. (b) is false since the numerical value of t is, like z , affected by outliers. t- procedures are generally OK to use for samples of size 40 or larger, but this is not what is meant by robust , so (d) is incorrect. (e) is not correct since the t- distribution is more variable than the standard normal. It becomes closer to z as sample size increases but is “as accurate” only in the limiting case.
2. The correct answer is (d). The alternative hypothesis, H 0 : μ = μ 0 , would require a negative value of z to be evidence against the null. Because the given value is positive, we conclude that the finding is in the wrong direction to support the alternative and, hence, is not going to be significant at any level.
3. The correct answer is (b). Because the data are paired, the appropriate t -test is a one-sample test for the mean of the difference scores. In this case, df = n – 1 = 8 – 1 = 7.
4. The correct answer is (a). The problem states that the teachers will record for comparison the number of students in each class who score above 80%. Since the enrollments differ in the two classes, we need to compare the proportion of students who score above 80% in each class. Thus the appropriate test is a two-proportion z -test. Note that, although it is not one of the choices, a chi-square test for homogeneity of proportions could also be used, since we are interested in whether the proportions of those getting above 80% are the same across the two populations.

Free-Response

1.

I . Let p 1 = the true proportion of students who admit to using marijuana in 2004.

Let p 2 = the true proportion of students who admit to using marijuana in 2007.

H 0 : p 1 = p 2 (or H 0 : p 1p 2 = 0; or H 0 : p 1p 2 ; or H 0 : p 1p 2 > 0).

H A : p 1 > p 2 (or H 0 : p 1 > p 2 ).

II . We will use a two-proportion z -test. The survey involved drawing random samples from independent populations. , 100(0.27) = 27, 100(1 – 0.27) = 73, 175(0.171) ≈ 30, and 175(1 – 0.171) ≈ 145 are all greater than 10, so the conditions for inference are present.

III .

(This problem can be done using 2-PropZTest in the STAT TESTS menu.)

IV . Since the P -value is quite small (a finding this extreme would occur only about 2.6% of the time if there had been no decrease in usage), we have evidence that the rate of marijuana use among students (at least among juniors and seniors) has decreased.

1. This is a paired study because the scores for each pair of twins are compared. Hence, it is a one-sample situation, and there are 26 pieces of data to be analyzed, which are the 26 difference scores between the twins. Hence, df = 26 – 1 = 25.
2. • I is not correct. A confidence interval, at least in AP Statistics, cannot be used in any one-sided hypothesis test—only two-sided tests.
• II is correct. A confidence interval constructed from a random sample that does not contain the hypothesized value of the parameter can be considered statistically significant evidence against the null hypothesis.
• III is not correct. The standard error for a confidence interval is based on the sample proportions is

The standard error for a significance test is based on the hypothesized population value is

• IV is correct.

4.

I . Let μ = the true mean score for all students taking the exam.

H 0: μ = 520 (or, H 0 : μ ≤ 520)

H A : μ > 520

II . We will use a one-sample t- test. We consider the 175 students to be a random sample of students taking the exam. The sample size is large, so the conditions for inference are present. (Note : Due to the large sample size, it is reasonable that the sample standard deviation is a good estimate of the population standard deviation. This means that you would receive credit on the AP exam for doing this problem as a z -test although a t -test is preferable.)

III . df = 175 – 1 = 174 ⇒ 0.05 < P -value < 0.10 (from Table B, with df = 100—always round down).

(Using the TI-83/84, P -value = tcdf(1.52,100,174)=0.065 . If we had used s = 96 as an estimate of s based on a large sample size, and used a z -test rather than a t -test, the P -value would have been 0.064. This is very close to the P -value determined using t . Remember that this is a t -test, but that the numerical outcome using a z -test is almost identical for large samples.)

IV . The P- value, which is greater than 0.05, is reasonably low but not low enough to provide strong evidence that the current class is superior in math ability as measured by the SAT.

5.

I . Let μ = the true average number of pages in the books Booker reads.

H 0 : μ ≤ 375.

H A : μ > 375.

II . We are going to use a one-sample t -test test at α = 0.05. The problem states that the sample is a random sample. A dotplot of the data shows good symmetry and no significant departures from normality (although the data do spread out quite a bit from the mean, there are no outliers):

The conditions for the t -test are present.

III .

(from Table B). (tcdf(1.48,100,12 ) gives P -value = 0.082, or you can use STAT TESTS T-Test .)

IV . Because P > .05, we cannot reject H 0 . We do not have strong evidence to back up Booker”s claim that the books he reads actually average more than 375 pages in length.

1. The data are paired by individual students, so we need to test the difference scores for the students rather than the means for each quiz. The differences are given in the following table.

I . Let μ d = the mean of the differences between the scores of students on Quiz 2 and Quiz 1.

H 0 : μ d = 0.

H A : μ d > 0.

II . This is a matched pairs t -test. That is, it is a one-sample t -test for a population mean. We assume that these are random samples from the populations of all students who took both quizzes. The significance level is α = 0.05.

A boxplot of the difference scores shows no significant departures from normality, so the conditions to use the one-sample t- test are present.

III .

(Using the TI-83/84, P -value = tcdf(2.33, 100,9)=0.022 , or you can use STAT TESTS T-Test .)

IV . Because P < 0.05, we reject the null hypothesis. The data provide evidence at the 0.05 level that the review was successful at improving student performance on the material.

7.

I . Let p = the true proportion of Wednesday night television viewers who are watching “I Want to Marry a Statistician.”

H 0 : p = 0.55.

H A : p < 0.55.

II . We want to use a one-proportion z-test at α = 0.01. 500(0.55) = 275 > 5 and 500(1 – 0.55) = 225 > 5. Thus, the conditions needed for this test have been met.

III .

(On the TI-83/84, normalcdf(-100,-1.80)=0.0359 , or you can use STAT TESTS 1-PropZTest. )

IV . Because P > 0.01, we do not have sufficient evidence to reject the null hypothesis. The evidence is insufficient to conclude at the 0.01 level that the proportion of viewers has dropped since the program was moved to a different night.

1. Let”s suppose that the Stats class constructed a 95% confidence interval for the true proportion of students who plan to vote for Harvey (we are assuming that this a random sample from the population of interest, and we note that both n and n (1 – ) are greater than 10). (as Harvey figured). Then a 95% confidence interval for the true proportion of votes Harvey can expect to get is . That is, we are 95% confident that between 35.5% and 71.2% of students plan to vote for Harvey. He may have a majority, but there is a lot of room between 35.5% and 50% for Harvey not to get the majority he thinks he has. (The argument is similar with a 90% CI: (0.384, 0.683); or with a 99% CI: (0.299, 0.768).)

9.

I . Let μ 1 = the true mean number of defects produced by machine A.

Let μ 2 = the true mean number of defects produced by machine B.

H 0 : μ 1μ 2 = 0.

H A : μ 1μ 2 ≠ 0.

II . We use a two-sample t -test for the difference between means. The conditions that need to be present for this procedure are given in the problem: both samples are simple random samples from independent, approximately normal populations.

III .

(from Table B).

(When read directly from Table B, t = –1.48 with df = 10 is between tail probabilities of 0.05 and 0.10. However, those are one-sided values and must be doubled for the two-sided alternative since we are interested in the probability of getting 1.48 standard deviations away from the mean in any direction. Using the TI-83/84, P -value = 2tcdf(-100,-1.48,10)= 0.170. Using the 2SampTTest from the STAT TESTS menu, P -value = 0.156 with df = 18.99.)

IV . The P -value is too large to be strong evidence against the null hypothesis that there is no difference between the machines. We do not have strong evidence that the types of machines actually differ in the number of defects produced.

1. Using a two-sample t -test, Steps I and II would not change. Step III would change to

based on df = min{30 – 1,30 – 1} = 29. Step IV would probably arrive at a different conclusion—reject the null because the P -value is small. Large sample sizes make it easier to detect statistically significant differences.

This P -value is quite low and provides evidence against the null and in favor of the alternative that security procedures actually detect less than the claimed percentage of banned objects.

12.

I . Let μ d = the mean of the differences between the scores of husbands and wives.

H 0 : μ d = 0.

H A : μ d < 0.

II . We are told to assume that σ = 1.77 (Note : This is a reasonable assumption, given the large sample size). This is a matched-pairs situation and we will use a one-sample z -test for a population mean. We assume that this is a random sample of married couples.

III.

(If you are bothered by using z rather than t —after all, we really don”t know σ D —note that for t = –2.84 based on df = 174, P -value = 0.0025, which is very close to the value obtained by using z and results in exactly the same conclusion.)

IV . Because P is very small, we reject H 0 . The data provide strong evidence that women have a stronger reaction to violence in videos than do men.

13.

I . Let p 1 = the true proportion of voters who plan to vote for Grumpy 4 weeks before the election.

Let p 2 = the true proportion of voters who plan to vote for Grumpy 2 weeks before the election.

H 0 : p 1p 2 = 0.

H A : p 1p 2 > 0.

II . We will use a two-proportion z- test for the difference between two population proportions. Both samples are random samples of the voting populations at the time.

Also,

All values are larger than 5, so the conditions needed for the two-proportion z- test are present.

III .

(From the TI-83/84, STAT TESTS 2-PropZTest yields P -value = 0.039.)

IV . Because P < 0.05, we can reject the null hypothesis. Candidate Dopey may have cause for celebration—there is evidence that support for candidate Grumpy is dropping.

14.

I . Let μ = the true mean popularity rating for “I Want to Marry a Statistician.”

H 0 : μ = 7.3.

H A : μ ≠ 7.3.

II . We will use a one-sample t confidence interval (at the direction of the problem—otherwise we would most likely have chosen to do a one-sample t significance test) at α = 0.05, which, for the two-sided test, is equivalent to a confidence level of 0.95 (C = 0.95). We will assume that the ratings are a random sample of the population of all ratings. The sample size is small, but we are told that the ratings are approximately normally distributed, so that the conditions necessary for the inference are present.

III . = 6.1, s = 2.7. For df = 12 – 1 = 11 and a 95% confidence interval, t * = 2.201 A 95% confidence interval for μ is given by:

(Note : A one-sample t test for these data yields P -value = 0.15.)

IV . Since 7.3 is in the interval (4.384, 7.816), we cannot reject H 0 at the 0.05 level of significance. We do not have good evidence that there has been a significantly significant change in the popularity rating of the show after its move to Wednesday night.

1. (a) df = min{23 – 1,27 – 1} = 22 ⇒ t * = 1.717.

(b) df = 22 ⇒ t * = 2.074.

(c) df = 23 + 27 –2 = 48 ⇒ t * = 1.684 (round down to 40 degrees of freedom in the table).

(d) df = 48 ⇒ t * = 2.021.

Solutions to Cumulative Review Problems

1. a.

(The exact binomial given by the TI-83/84 is 1-binomcdf(250,0.6,160) = 0.087 . If you happen to be familiar with using a continuity correction (and you don”t really need to be) for the normal approximation, normalcdf (160.5,1000,150,7.75) = 0.088 , which is closer to the exact binomial.)

1. . (On the TI-83/84, this is equivalent to 1-binomcdf(5,0.3,2).)
2. P (the worker is an administrator | the worker is female)
3. A confidence interval is justified because we are dealing with a random sample from an approximately normally distributed population. df = 25 – 1 = 24 ⇒ t * = 1.711.

We are 90% confident that the true mean number of mg per cigarette for this type of cigarette is between 1.59 mg and 2.11 mg.

CHAPTER 12

Inference for Means and Proportions

1. In a study to see if special exercises help babies learn to walk sooner, researchers randomly assigned 30 babies to experience the exercise regimen starting at the age of three months; 30 other babies were used as a control group and did not experience the exercises. The researchers recorded the age at which each baby first walked. A t -test for the difference in means yielded a P -value of 0.56. What does this P -value mean in context?

(A) There is a 56% chance that the exercises helped babies to walk sooner.

(B) There is a 56% chance that there is no difference between babies who do the exercises and those who do not.

(C) The exercises caused a 56% improvement in the time it takes for babies to walk.

(D) If experiencing the exercises helps babies walk sooner, there is a 56% chance of seeing a difference as large as or larger than the one seen in this experiment.

(E) If there is no difference between babies who experience the exercises and those who do not, there is a 56% chance of seeing a difference as large as or larger than the one seen in this experiment.

1. Seventy-five percent of crashes involving teen drivers were caused by errors on the part of the teen driver. A simple random sample of 80 crashes involving teen drivers is selected. The sampling distribution of the sample proportion has a mean of 0.75 and a standard deviation of

(A) 0.0054.

(B) 0.0484.

(C) 0.0949.

(D) 0.4330.

(E) 3.8730.

1. A nutritionist was studying the effect of a particular diet on sleep loss in patients who had reported difficulty sleeping. She conducted a test with the following hypotheses: H0 : μ N = μ C where μ N is the mean number of hours of sleep for those on the new diet, and μ C is the mean number of hours of sleep for those in the control group.

H A : μ N > μ C

She conducted the experiment and performed a significance test, which resulted in a P -value of 0.77. Which statement describes the correct conclusion?

(A) Because the P -value is greater than 0.5, there is evidence that μ N > μ C .

(B) Because the P- value is low, there is evidence that μ N = μ C .

(C) Because the P- value is high, there is evidence that μ N = μ C .

(D) Because the P- value is low, there is insufficient evidence to say μ N = μ C .

(E) Because the P- value is high, there is insufficient evidence to say μ N > μ C .

1. A company is planning an experiment to determine whether a new insect repellent formula is more effective than its old formula. The company plans to randomly select one arm on each volunteer to be sprayed with the new formula, and the other arm will be sprayed with the old formula. Each arm will be put into an aquarium full of mosquitos for five minutes. The researchers will count the number of bites on each arm. Which of the following would be the most appropriate inference technique?

(A) A z -test for a difference in proportions

(B) A t -test for a difference in means

(C) A confidence interval for a difference in proportions

(D) A one-sample t -test for matched pairs

(E) Two confidence intervals, one for each insecticide formula

1. A polling organization conducted a survey of 850 randomly selected U.S. adults one year ago and asked whether they were happy with the state of their community. Seventy-eight percent said yes. This year the organization polled 925 randomly selected U.S. adults, and 81% answered yes to the same question. The pollsters will conduct a significance test to see if they have evidence that the percentage of U.S. adults who are happy with their community has changed. What should the standard deviation used in the denominator of the test statistic z be?

(A)

(B)

(C)

(D)

(E)