Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For the following pairs, indicate which do not comply with the rules for setting up hypotheses, and explain why: a. \(H_{0}: \mu=15, H_{a}: \mu=15\) b. \(H_{0}: \pi=.4, H_{a}: \pi>.6\) c. \(H_{0}: \mu=123, H_{a}: \mu<123\) d. \(H_{0}: \mu=123, H_{a}: \mu=125\) e. \(H_{0}: p=.1, H_{a}: p=125\)

Short Answer

Expert verified
The pairs which do not comply with the rules for setting up hypotheses are a. \(H_{0}: \mu=15, H_{a}: \mu=15\), b. \(H_{0}: \pi=.4, H_{a}: \pi>.6\), d. \(H_{0}: \mu=123, H_{a}: \mu=125\) and e. \(H_{0}: p=.1, H_{a}: p=125\). The reasons include - both hypotheses being the same (pair a), not covering all possible values of the parameter (pair b), hypotheses not being mutually exclusive (pair d), and incorrect parameter value (pair e).

Step by step solution

01

Analyzing first pair

In pair a. \(H_{0}: \mu=15, H_{a}: \mu=15\), both hypotheses are the same. This violates the rule that \(H_{0}\) and \(H_{a}\) are mutually exclusive. Hence, this pair doesn't comply with the rules.
02

Analyzing second pair

In pair b. \(H_{0}: \pi=.4, H_{a}: \pi>.6\), the sum of \(H_{0}\) and \(H_{a}\) doesn’t cover values between .4 and .6 which violates one of the rules, hence, this pair also doesn't comply.
03

Analyzing third pair

In pair c. \(H_{0}: \mu=123, H_{a}: \mu<123\), the hypotheses are mutually exclusive and the equality sign is in the correct hypothesis. So, this pair complies with the rules.
04

Analyzing fourth pair

In pair d. \(H_{0}: \mu=123, H_{a}: \mu=125\), we see that these two hypotheses are not mutually exclusive which violates one of the rules. Hence, this pair doesn't comply with the rules.
05

Analyzing fifth pair

In pair e. \(H_{0}: p=.1, H_{a}: p=125\), the alternative hypothesis clearly doesn’t make sense because a probability should lie between 0 and 1 inclusive, hence this pair does not comply with the rules.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is fundamental in hypothesis testing, an integral part of statistics. The null hypothesis, denoted as \(H_{0}\), typically represents a statement of no effect or no difference. It is the hypothesis that researchers aim to test against the alternative hypothesis. The alternative hypothesis, expressed as \(H_{a}\), proposes what we suspect might actually be true or what we wish to test for. For example, if we want to test if a new drug is more effective than the current standard, the null hypothesis would say the new drug has the same effectiveness as the current one, while the alternative would claim it is more effective.

The two hypotheses must be framed so they are mutually exclusive and exhaustive; that means they cannot overlap and must cover all possible outcomes. An error in setting these can invalidate the entire test. For instance, if the null says a coin is fair (\(p = 0.5\)), the alternative should be that the coin is not fair (\(p eq 0.5\)), thus covering all other possibilities, such as \(p > 0.5\) or \(p < 0.5\). It's essential to state these hypotheses correctly and clearly as they set the stage for all subsequent steps in hypothesis testing.
Mutually Exclusive Hypotheses
The concept of 'mutually exclusive hypotheses' can sometimes be a tricky one for students. Essentially, hypotheses are mutually exclusive if they do not overlap and cannot both be true at the same time. In any hypothesis test, the null and alternative hypotheses should be mutually exclusive to ensure a clear decision can be made from the test results. In the exercise provided, the fourth pair \(H_{0}: \mu=123, H_{a}: \mu=125\) is an example of hypotheses that are not mutually exclusive. This is incorrect because they do not cover all possibilities and both could potentially be true at the same time. If the population mean \(\mu\) was actually 124, both \(H_{0}\) and \(H_{a}\) would be incorrect. Instead, the alternative hypothesis should be something like \(H_{a}: \mu eq 123\) to contrast \(H_{0}\) and cover all other values except 123.

When hypotheses are mutually exclusive, only one can be true. This is vital for the validity of the test, for interpreting results correctly, and for making sound conclusions. If this criterion isn't met, the entire hypothesis testing process can lead to ambiguous or misleading outcomes, rendering the analysis scientifically weak or useless.
Probability in Statistics
Probability plays a central role in statistics, and it forms the backbone of hypothesis testing. It is the measure of the likelihood that an event will occur, expressed as a number between 0 (the event will not occur) and 1 (the event will occur for sure). For instance, when a coin is flipped, the probability of landing a head or a tail is each \(0.5\) if the coin is fair. This is because there are only two equally likely outcomes.

In hypothesis testing, we use probability to make conclusions about the population based on sample data. We consider the probability of observing our sample data, or something more extreme, if the null hypothesis were true. This p-value helps us decide whether to reject or fail to reject the null hypothesis. If the p-value is below a certain threshold (commonly \(0.05\)), we reject the null hypothesis in favor of the alternative.

The fifth pair in the exercise, \(H_{0}: p=.1, H_{a}: p=125\), misunderstands the concept of probability since probabilities cannot exceed 1. A correct alternative hypothesis might state \(H_{a}: p eq .1\) or \(H_{a}: p > .1\) depending on the context, reflecting probabilities that are consistent with defined boundaries (0 to 1). Grasping the nature of probability is crucial because it helps us quantify uncertainty and make informed decisions based on statistical evidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mu\) denote the true average diameter for bearings of a certain type. A test of \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) will be based on a sample of \(n\) bearings. The diameter distribution is believed to be normal. Determine the value of \(\beta\) in each of the following cases: a. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.52\) b. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.48\) c. \(n=15, \alpha=.01, \sigma=0.02, \mu=0.52\) d. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.54\) e. \(n=15, \alpha=.05, \sigma=0.04, \mu=0.54\) f. \(n=20, \alpha=.05, \sigma=0.04, \mu=0.54\) g. Is the way in which \(\beta\) changes as \(n, \alpha, \sigma\), and \(\mu\) vary consistent with your intuition? Explain.

A survey of teenagers and parents in Canada conducted by the polling organization Ipsos ("Untangling the Web: The Facts About Kids and the Internet," January \(25 .\) 2006) included questions about Internet use. It was reported that for a sample of 534 randomly selected teens, the mean number of hours per week spent online was \(14.6\) and the standard deviation was \(11.6\). a. What does the large standard deviation, \(11.6\) hours, tell you about the distribution of online times for this sample of teens? b. Do the sample data provide convincing evidence that the mean number of hours that teens spend online is greater than 10 hours per week?

An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 4 min. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was \(1.4\) min. Is there convincing evidence that the average attention span for teenage boys is less than 5 min? Test the relevant hypotheses using \(\alpha=.01\).

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed 100 \(\mathrm{lb} / \mathrm{in} .^{2}\). The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

Duck hunting in populated areas faces opposition on the basis of safety and environmental issues. The San Luis Obispo Telegram-Tribune (June 18,1991 ) reported the results of a survey to assess public opinion regarding duck hunting on Morro Bay (located along the central coast of California). A random sample of 750 local residents included 560 who strongly opposed hunting on the bay. Does this sample provide sufficient evidence to conclude that the majority of local residents oppose hunting on Morro Bay? Test the relevant hypotheses using \(\alpha=.01\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free