Review the following concepts if necessary:

- Significance Testing for AP Statistics
- Inference for a Single Population Mean for AP Statistics
- Inference for the Difference Between Two Population Means for AP Statistics
- Inference for a Single Population Proportion for AP Statistics
- Inference for the Difference Between Two Population Proportions for AP Statistics

### Rapid Review

- A researcher reports that a finding of is significant at the 0.05 level of significance. What is the meaning of this statement?
- Let μ
_{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_{1}– μ_{2}= 0. A large sample significance test finds*P*= 0.04. What conclusion, in the context of the problem, do you report if- α = 0.05?
- α = 0.01?

- 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. - 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. - True–False: In a hypothesis test concerning a single mean, we can use either
*z*-procedures or*t*-procedures as long as the sample size is at least 20. - We are going to conduct a two-sided significance test for a population proportion. The null hypothesis is
*H*_{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- you are constructing a confidence interval for the true population proportion?
- you are doing a significance test for the null hypothesis?

- For a confidence interval, you use the value of in the standard error. Hence, .
- For a significance test, you use the hypothesized value of
*p*. Hence, - For the following data,
- justify the use of a two-proportion
*z*-test for*H*_{0}:*p*_{1}–*p*_{2}= 0. - what is the value of the test statistic for
*H*_{0}:*p*_{1}–*p*_{2}= 0? - what is the
*P*-value of the test statistic for the two-sided alternative?

- justify the use of a two-proportion
*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).- 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. - Although it may be difficult to justify, there are conditions under which you can
*pool*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:* 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.

*Answer:*

*Answer:* 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.

*Answer:*

.

.

*Answer:*

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

.

.

*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.

*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.

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