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Hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu=15\) vs \(H_{a}: \mu \neq 15\) (a) \(95 \%\) confidence interval for \(\mu: \quad 13.9\) to 16.2 (b) \(95 \%\) confidence interval for \(\mu: \quad 12.7\) to 14.8 (c) \(90 \%\) confidence interval for \(\mu: \quad 13.5\) to 16.5

Short Answer

Expert verified
For samples (a) and (c), we do not reject the null hypothesis at 5% and 10% significance levels respectively while for sample (b), we reject the null hypothesis at a 5% significance level.

Step by step solution

01

Analyze the first confidence interval

For part (a), the confidence interval ranges from 13.9 to 16.2. Since the hypothesized population mean of 15 lies within this interval, we do not reject the null hypothesis at a 5% significance level (we are using a 95% confidence interval).
02

Analyze the second confidence interval

For part (b), the confidence interval ranges from 12.7 to 14.8. The hypothesized population mean of 15 does not fall within this interval, so we reject the null hypothesis at a 5% significance level.
03

Analyze the third confidence interval

For part (c), the confidence interval ranges from 13.5 to 16.5. Since the hypothesized population mean of 15 is within this range, we do not reject the null hypothesis at a 10% significance level (because we are using a 90% confidence interval).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval is a range of values, derived from the sample data, that is likely to contain the population parameter of interest. When we calculate, for example, a 95% confidence interval, we are saying that we are 95% confident that the true population mean falls within this range.

In the context of the exercise, the confidence intervals are given for the population mean \( \mu \). When the sample mean falls within the confidence interval, such as in parts (a) and (c), we do not have sufficient evidence to reject the null hypothesis. This is because the hypothesized value of \( \mu=15 \) is within the range that we're 95% or 90% confident contains the true mean.

However, for part (b), the hypothesized mean does not fall within the 95% confidence interval. This suggests that the true mean is likely different from 15, and we would reject the null hypothesis in this scenario. Confidence intervals are an essential part of statistical hypothesis testing because they provide a range of plausible values for the parameter, allowing for a decision on the null hypothesis based on the data at hand.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or no difference, and it generally represents a skeptical perspective or a claim to be tested. In hypothesis testing, we seek to determine whether the evidence suggests that we should reject this null hypothesis in favor of an alternative hypothesis, denoted as \( H_a \).

For instance, in the given exercise, \( H_0: \mu=15 \) asserts that the population mean is 15. The alternative hypothesis \( H_a: \mu eq 15 \) posits that the population mean is not 15. The null hypothesis is the starting assumption for the test, and the hypothesis testing procedure examines whether the data collected provides enough evidence to conclude if the null hypothesis can be rejected or not.

As we can see from the solutions, a confidence interval that does not include the value stated in the null hypothesis (as in part (b)) is an indicator that the null hypothesis may not hold. Conversely, when the confidence interval includes the null hypothesis value, we lack evidence to reject it (as seen in parts (a) and (c)).
Significance Level
The significance level, often denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. It represents the researcher's tolerance for such errors and is a critical value in hypothesis testing that helps determine the threshold for rejecting the null hypothesis.

Common significance levels are 5% (0.05), 1% (0.01), or 10% (0.10), which corresponds inversely to 95%, 99%, and 90% confidence levels, respectively. Demonstrated in our exercise, when we reject the null hypothesis for part (b), it's because the 95% confidence interval does not include the hypothesized mean of 15, thus surpassing the 5% significance level criterion for rejection.

Alternatively, for parts (a) and (c), the hypothesized mean lies within the confidence intervals, indicating that we do not have significant evidence at the 5% and 10% levels, respectively, to reject the null hypothesis. Deciding on the appropriate significance level is a crucial step in the design of an experiment or study as it can influence the conclusions drawn from the statistical test.

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Most popular questions from this chapter

Does the airline you choose affect when you'll arrive at your destination? The dataset DecemberFlights contains the difference between actual and scheduled arrival time from 1000 randomly sampled December flights for two of the major North American airlines, Delta Air Lines and United Air Lines. A negative difference indicates a flight arrived early. We are interested in testing whether the average difference between actual and scheduled arrival time is different between the two airlines. (a) Define any relevant parameter(s) and state the null and alternative hypotheses. (b) Find the sample mean of each group, and calculate the difference in sample means. (c) Use StatKey or other technology to create a randomization distribution and find the p-value. (d) At a significance level of \(\alpha=0.01\), what is the conclusion of the test? Interpret the conclusion in context.

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: \(H_{0}: p=0.6\) vs \(H_{a}: p>0.6\) Sample data: \(\hat{p}=52 / 80=0.65\) with \(n=80\)

The Ignorance Surveys were conducted in 2013 using random sampling methods in four different countries under the leadership of Hans Rosling, a Swedish statistician and international health advocate. The survey questions were designed to assess the ignorance of the public to global population trends. The survey was not just designed to measure ignorance (no information), but if preconceived notions can lead to more wrong answers than would be expected by random guessing. One question asked, "In the last 20 years the proportion of the world population living in extreme poverty has \(\ldots, "\) and three choices were provided: 1) "almost doubled" 2) "remained more or less the same," and 3) "almost halved." Of 1005 US respondents, just \(5 \%\) gave the correct answer: "almost halved." 34 We would like to test if the percent of correct choices is significantly different than what would be expected if the participants were just randomly guessing between the three choices. (a) What are the null and alternative hypotheses? (b) Using StatKey or other technology, construct a randomization distribution and compute the p-value. (c) State the conclusion in context.

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 \(n=50\)

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) using \(\hat{p}=0.55\) with each of the following sample sizes: (a) \(\hat{p}=55 / 100=0.55\) (b) \(\hat{p}=275 / 500=0.55\) (c) \(\hat{p}=550 / 1000=0.55\)

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