/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Which of the following \(95 \%\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following \(95 \%\) confidence intervals would lead us to reject \(H_{0}: p=0.30\) in favor of \(H_{a}: p \neq 0.30\) at the \(5 \%\) significance level? (a) (0.19,0.27) (c) (0.27,0.31) (e) None of these (b) (0.24,0.30) (d) (0.29,0.38)

Short Answer

Expert verified
Interval (a) leads to rejecting H_0 .

Step by step solution

01

Understanding Confidence Intervals and Hypotheses

First, understand that a 95% confidence interval (CI) is used to estimate a population parameter and check if that interval contains the hypothesized parameter. The null hypothesis ( H_0 ) states that the parameter is 0.30, and the alternative hypothesis ( H_a ) suggests it's different from 0.30.
02

Identify which Intervals Exclude 0.30

Evaluate each confidence interval to see if the value 0.30 is included. If 0.30 is within the interval, we fail to reject H_0 . If 0.30 is not within the interval, we reject H_0 in favor of H_a .
03

Check Individual Intervals

- Interval (a) (0.19, 0.27): 0.30 is not included. - Interval (b) (0.24, 0.30): 0.30 is included. - Interval (c) (0.27, 0.31): 0.30 is included. - Interval (d) (0.29, 0.38): 0.30 is included.
04

Determine Correct Answer

Only interval (a) (0.19, 0.27) does not contain 0.30, leading us to reject H_0 . The other intervals include 0.30, implying we cannot reject H_0 for them.

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 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. It starts with forming a hypothesis about a population parameter, such as a proportion or a mean. Typically, hypothesis testing involves two hypotheses:
  • The null hypothesis ( $H_0$ ) - a statement of no effect or no difference. It's a hypothesis one tries to reject.
  • The alternative hypothesis ( $H_a$ ) - a statement that indicates the presence of an effect or difference.
The process involves calculating a test statistic, which is a function of the sample data, and comparing it to a threshold value derived from a significance level. If the test statistic exceeds the threshold, the null hypothesis is rejected in favor of the alternative hypothesis. It’s a structured way to test if your assumptions align with reality by using sample data.
Significance Level
Significance level, often denoted as \(\alpha\), is a threshold set by the researcher which determines how strong the evidence must be to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. Common significance levels are 0.05, 0.01, and 0.10.

For example, a significance level of 0.05 indicates a 5% risk of concluding that a difference exists when there is no actual difference. Lower significance levels indicate stronger evidence required to reject the null hypothesis.
  • If \(\alpha = 0.05\), findings are significant at the 5% level.
  • If the test result is 0.05 or less, the null hypothesis is rejected.
Setting a significance level helps control Type I errors, where one incorrectly rejects the true null hypothesis.
Null Hypothesis
The null hypothesis, denoted as $H_0$ , is a key concept in hypothesis testing. It is the hypothesis that there is no effect or no difference, and it serves as the starting point for statistical testing. In many cases, it represents a statement of "no change" or "status quo."
  • For instance, if you're testing a coin flip, $H_0$ might be "the coin is fair," meaning it lands on heads 50% of the time.
  • The goal of hypothesis testing is to assess whether the sample data provides enough evidence to reject $H_0$ in favor of an alternative hypothesis.
A crucial step in hypothesis testing is to avoid bias towards any outcome. Therefore, $H_0$ is presumed true until evidence indicates otherwise. Rejecting the null hypothesis suggests that there might be an effect present.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_a\) or \(H_1\), proposes an opposite effect or a change from the null hypothesis. It represents what you aim to support in your study. In our exercise, while the null hypothesis asserts \((H_0: p = 0.30)\), the alternative hypothesis states \((H_a: p eq 0.30)\), implying that the population proportion is different from 0.30.
  • The alternative hypothesis needs strong statistical evidence to be supported over the null hypothesis.
  • There are two types: a one-sided hypothesis (greater or less) or a two-sided hypothesis (not equal).
In testing terms, supporting the alternative hypothesis indicates that your sample shows significant differences that lead you away from the null hypothesis, making it useful in real-world decision-making processes.

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

In the sample, \(\hat{p}=158 / 300=0.527 .\) The resulting \(P\) -value is 0.18 . What is the correct interpretation of this \(P\) -value? (a) Only \(18 \%\) of the city residents support the tax increase. (b) There is an \(18 \%\) chance that the majority of residents supports the tax increase. (c) Assuming that \(50 \%\) of residents support the tax increase, there is an \(18 \%\) probability that the sample proportion would be 0.527 or higher by chance alone. (d) Assuming that more than \(50 \%\) of residents support the tax increase, there is an \(18 \%\) probability that the sample proportion would be 0.527 or higher by chance alone. (e) Assuming that \(50 \%\) of residents support the tax increase, there is an \(18 \%\) chance that the null hypothesis is true by chance alone.

The composition of the earth's atmosphere may have changed over time. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on 9 specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen: \(^{20}\) $$ \begin{array}{llllllll} 63.4 & 65.0 & 64.4 & 63.3 & 54.8 & 64.5 & 60.8 & 49.1 & 51.0 \end{array} $$ Explain why we should not carry out a one-sample \(t\) test in this setting.

How well materials conduct heat matters when designing houses, for example. Conductivity is measured in terms of watts of heat power transmitted per square meter of surface per degree Celsius of temperature difference on the two sides of the material. In these units, glass has conductivity about \(1 .\) The National Institute of Standards and Technology provides exact data on properties of materials. Here are measurements of the heat conductivity of 11 randomly selected pieces of a particular type of glass: \({ }^{22}\) \(\begin{array}{llllllllllll}1.11 & 1.07 & 1.11 & 1.07 & 1.12 & 1.08 & 1.08 & 1.18 & 1.18 & 1.18 & 1.12\end{array}\) (a) Is there convincing evidence that the mean conductivity of this type of glass is greater than \(1 ?\) (b) Given your conclusion in part (a), which kind of mistake-a Type I error or a Type II error - could you have made? Explain what this mistake would mean in context.

Multiple choice: Select the best answer for Exercises 95 to 102 . The reason we use \(t\) procedures instead of \(z\) procedures when carrying out a test about a population mean is that (a) \(z\) requires that the sample size be large. (b) \(z\) requires that you know the population standard deviation \(\sigma\). (c) \(z\) requires that the data come from a random sample or randomized experiment. (d) \(z\) requires that the population distribution be perfectly Normal. (e) \(z\) can only be used for proportions.

In Exercises 7 to 10, explain what's wrong with the stated hypotheses. Then give correct hypotheses. A change is made that should improve student satisfaction with the parking situation at a local high school. Right now, \(37 \%\) of students approve of the parking that's provided. The null hypothesis \(H_{0}: p>0.37\) is tested against the alternative \(H_{a}: p=0.37\)

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.