AP Stats Review of Significance Tests

• Since our p-value of 0.0391 is more than 5%, we reject the null hypothesis.  We have convincing evidence that the mean copper content of your bottles is greater than 1.3 mg/l.

• Since our p-value of 0.0391 is more than 5%, we fail to reject the null hypothesis.  We do not have convincing evidence that the mean copper content of your bottles is greater than 1.3 mg/l.

• Since our p-value of 0.0391 is less than 5%, we reject the null hypothesis.  We have convincing evidence that the mean copper content of your bottles is greater than 1.3 mg/l.

• Since our p-value of 0.0391 is less than 5%, we fail to reject the null hypothesis.  We do not have convincing evidence that the mean copper content of your bottles is greater than 1.3 mg/l.

Simon reads a newspaper report claiming that 12% of all adults in the US are left-handed. He wonders if this figure holds true at his college. Simon chooses an SRS of 100 students and 16 of them are left-handed. He performs a test of Ho: p = 0.12 versus Ha: =/= 0.12. The test yields a p-value of 0.2184. What conclusion would you make for a significance level of α= 0.10?

• Since the p-value of 0.2184 >α, we reject the null hypothesis. We have convincing evidence that the proportion of lefties at his college is different from 12%.

• Since the p-value of 0.2184 >α, we fail to reject the null hypothesis. We do not have convincing evidence that the proportion of lefties at his college is different from 12%.

• Since the p-value of 0.2184 <α, we reject the null hypothesis. We have convincing evidence that the proportion of lefties at his college is different from 12%.

• Since the p-value of 0.2184 <α, we fail to reject the null hypothesis. We do not have convincing evidence that the proportion of lefties at his college is different from 12%.

Simon reads a newspaper report claiming that 12% of all adults in the US are left-handed. He wonders if this figure holds true at his college. Simon chooses an SRS of 100 students and 16 of them are left-handed. He performs a test of Ho: p = 0.12 versus Ha: =/= 0.12. The test yields a p-value of 0.2184. What conclusion would you make for a significance level of α = 0.05?

• Since the p-value of 0.2184 > α, we reject the null hypothesis. We have convincing evidence that the proportion of lefties at his college is different from 12%.

• Since the p-value of 0.2184 > α, we fail to reject the null hypothesis. We do not have convincing evidence that the proportion of lefties at his college is different from 12%.

• Since the p-value of 0.2184 < α, we reject the null hypothesis. We have convincing evidence that the proportion of lefties at his college is different from 12%.

• Since the p-value of 0.2184 < α, we fail to reject the null hypothesis. We do not have convincing evidence that the proportion of lefties at his college is different from 12%.

Which of the following best describes a Type I error?

• The null is true, but we mistakenly reject it.

• The null is false and we reject it.

• The null is false, but we fail to reject it.

• The null is true but we fail to reject it.

• The null is true, and we fail to reject it.

To test a claim about a mean, when the population standard deviation is unknown we use:

• z procedures

• Pythagorean Theorem

• t procedures

• np > 10 and n(1-p) > 10

Margin of error equals:

• Critical Value ∗ standard Error

• z*

• 1.96

• Standard Error

Which of the following best describes a type II error?

• The null is true, but we mistakenly reject it.

• The null is false and we reject it.

• The null is false, but we fail to reject it.

• The null is true but we fail to reject it.

• Te null is false, and we reject it.

When conducting a significance test for the difference in proportions, why do we pool the data when finding standard error?

• Because we like to swim

• Because we assume p1 = p2 in Ho

• Because the sample sizes are always equal

• To be safe and make sure we don't underestimate the standard error

• Because we don't the standard deviation.

A chi-square test for homogeneity was conducted to investigate whether the four high schools in a school district have different absentee rates for each of four grade levels. The chi-square test statistic and p-value of the test were 19.02 and 0.025, respectively. Which of the following is the correct interpretation of the p-value in the context of the test?

• Assuming that each high school has the same absentee rate for each grade level, there is a 2.5 percent chance of finding a test statistic 19.02 or smaller.

• Assuming that each high school has the same absentee rate for each grade level, there is a 2.5 percent chance of finding a test statistic 19.02 or larger.

• Assuming that each high school has a different absentee rate for each grade level, there is a 2.5 percent chance of finding a test statistic 19.02 or larger.

• There is a 2.5 percent chance that the absentee rate for each grade level at the four schools is the same.

• There is a 2.5 percent chance that the absentee rate for each grade level at the four schools is different.

Answer explanation

The null hypothesis is that there is no difference in the absentee rates for the four high schools. Under the assumption that the null hypothesis is true, a p-value of 0.025 means that there is a 2.5 percent probability of obtaining a test statistic value as extreme as or more extreme than a test statistic value of 19.02.

In 2010, a medical research group reported the results of an experiment to evaluate the effectiveness of acupuncture to treat a chronic intestinal condition. A group of volunteers with the chronic intestinal condition agreed to participate in the experiment and be randomly assigned to either a true acupuncture treatment or a placebo treatment. The placebo treatment mimicked the application of acupuncture, but no needle penetrated the skin. Random assignment resulted in 78 subjects receiving acupuncture and 75 subjects receiving the placebo treatment. After receiving 6 treatments over the course of 3 weeks, patients were asked to report whether they had experienced a reduction in the chronic intestinal condition. The table summarizes the data from the study, with expected cell counts in parentheses. Which of the following is true about the chi-square test for homogeneity?

• The number of subjects randomly assigned to each treatment is not the same; therefore, it is not appropriate to use a chi-square test for homogeneity across treatment groups.

• Volunteers do not constitute a random sample from the population of all patients with the chronic intestinal condition; therefore, it is not appropriate to use a chi-square test for homogeneity across treatment groups.

• Volunteers with the chronic intestinal condition were randomly assigned to each treatment, so the independence condition has been met.

• Not all of the observed cell counts are large enough to satisfy the conditions for applying the chi-square test of homogeneity.

• Not all of the expected cell counts are large enough to satisfy the conditions for applying the chi-square test for homogeneity.

Answer explanation

Since this was an experiment, treatments had to be randomly assigned, which satisfies the condition for independence. Also, because the experiment consists of volunteers and there was not a sample selected from the population of interest, it is not necessary to check that the sample size is less than or equal to 10 percent of the population size.