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Walking to school A recent report claimed that \(13 \%\) of students typically walk to school. \({ }^{10}\) DeAnna thinks that the proportion is higher than 0.13 at her large elementary school, so she surveys a random sample of 100 students to find out.

Short Answer

Expert verified
Without the exact sample data or significance level, we can't conclude if the proportion is higher than 0.13.

Step by step solution

01

Identify the Hypotheses

We want to determine if the proportion of students who walk to school is greater than 0.13. The null hypothesis, denoted as \(H_0\), is that the true proportion \(p = 0.13\). The alternative hypothesis, \(H_a\), is that \(p > 0.13\).
02

Collect and Analyze the Sample Data

DeAnna surveys a random sample of 100 students. Let's assume she finds that 15 students in her sample walk to school. The sample proportion \(\hat{p}\) is \(\frac{15}{100} = 0.15\).
03

Calculate the Test Statistic

Use the formula for the test statistic for a proportion: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \(p_0 = 0.13\), \(\hat{p} = 0.15\), and \(n=100\). Therefore, \( z = \frac{0.15 - 0.13}{\sqrt{\frac{0.13 \times 0.87}{100}}} \).
04

Find the P-value

The P-value corresponds to the probability of observing a test statistic as extreme as, or more extreme than, the one observed, under the null hypothesis. Look up the calculated z-value in a standard normal distribution table to find this probability.
05

Make a Conclusion

If the P-value is less than the significance level (commonly \(0.05\)), we reject the null hypothesis. Suppose if the P-value was found to be 0.08, we would not reject the null hypothesis at xx level of significance. Thus, there's not enough evidence to say the proportion is greater than 0.13.

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Key Concepts

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

Proportion
In statistics, a **proportion** refers to a part of the whole, expressed as a fraction, percentage, or decimal. It represents the ratio of a subset to the entire population.
For example, if out of 100 students surveyed, 15 walk to school, the proportion is \ \( \frac{15}{100} = 0.15 \ \).

When conducting surveys or experiments, calculating proportions allows us to understand and compare different subsets of data within a population. This is particularly useful in hypothesis testing to determine if observed sample proportions differ from specified values such as claims or expectations.

  • Proportions are used to draw conclusions about a population based on a sample.
  • They provide a simple way to report the prevalence of a particular characteristic within a group.
Null Hypothesis
The **Null Hypothesis** (often represented as \( H_0 \)) is a fundamental concept in hypothesis testing. It states that there is no effect or no difference, and it serves as the starting point for statistical testing.
For DeAnna's study, the null hypothesis is \( p = 0.13 \), meaning the proportion of students who walk to school is 13\% or less.

The null hypothesis is usually set up to be tested against the alternative hypothesis, \( H_a \), which suggests that the actual effect or difference exists.

Key aspects of \( H_0 \):
  • It is assumed true until evidence suggests otherwise.
  • Serves as a baseline for comparison to observe sample data.
  • Its rejection implies support for the alternative hypothesis.
Understanding the null hypothesis is crucial because it guides the direction and interpretation of statistical tests.
Test Statistic
The **Test Statistic** is a value calculated from sample data that is used to evaluate the null hypothesis. It quantifies the degree of agreement between the observed data and \( H_0 \).
In proportion hypothesis testing, the test statistic is often expressed as a \( z \)-score.

For DeAnna's survey, the formula for the test statistic is:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]
where:
\( \hat{p} \) = sample proportion (0.15),
\( p_0 \) = hypothesized population proportion (0.13),
\( n \) = sample size (100).

A higher absolute \( z \)-score indicates that the observed proportion deviate more from the hypothesized proportion.

  • The test statistic helps to determine whether to reject \( H_0 \).
  • A large \( |z| \)-value suggests that the sample proportion is significantly different from the hypothesized proportion.
P-value
The **P-value** measures the probability of obtaining a test statistic as extreme as the observed one, assuming the null hypothesis is true.
It helps determine the statistical significance of the test.

In hypothesis testing, a small P-value indicates strong evidence against \( H_0 \), prompting researchers to reject the null hypothesis. Conversely, a large P-value suggests there is not enough evidence to refute \( H_0 \).

For instance, in DeAnna's test, if the P-value is 0.08, this means there is an 8\% chance of observing a sample statistic as extreme as the one found, if \( H_0 \) were true.

  • P-values help assess the strength of evidence against the null hypothesis.
  • Common significance levels for making decisions are 0.05, 0.01, and 0.1.
Thus, P-values play an essential role in deciding whether the results of an hypothesis test are statistically significant.

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Most popular questions from this chapter

After once again losing a football game to the archrival, a college's alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of living alumni are in favor of firing the coach. The appropriate test statistic is (a) \(z=\frac{0.64-0.5}{\sqrt{\frac{0.64(0.36)}{100}}}\) (b) \(t=\frac{0.64-0.5}{\sqrt{\frac{0.64(0.36)}{100}}}\) (c) \(z=\frac{0.64-0.5}{\sqrt{\frac{0.5(0.5)}{100}}}\) (d) \(z=\frac{0.64-0.5}{\sqrt{\frac{0.64(0.36)}{64}}}\) (e) \(z=\frac{0.5-0.64}{\sqrt{\frac{0.5(0.5)}{100}}}\)

A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter \(\mu=4.2\) inches and standard deviation \(\sigma=0.05\) inches. The CD manufacturer decides to produce CDs with mean diameter \(\mu=4\) inches. Their diameters follow a Normal distribution with \(\sigma=0.1\) inches. (a) Let \(X=\) the diameter of a randomly selected \(\mathrm{CD}\) and \(Y=\) the diameter of a randomly selected case. Describe the shape, center, and spread of the distribution of the random variable \(X-Y\). What is the importance of this random variable to the CD manufacturer? (b) Compute the probability that a randomly selected CD will fit inside a randomly selected case. (c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases.

The \(z\) statistic for a test of \(H_{0}: p=0.4\) versus \(H_{a}: p \neq 0.4\) is \(z=2.43 .\) This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at \(\alpha=0.01\) but not at \(\alpha=0.05\). (d) significant at both \(\alpha=0.05\) and \(\alpha=0.01\). (e) inconclusive because we don't know the value of \(\hat{p}\).

You manufacture and sell a liquid product whose electrical conductivity is supposed to be \(5 .\) You plan to make six measurements of the conductivity of each lot of product. If the product meets specifications, the mean of many measurements will be \(5 .\) You will therefore test $$ \begin{array}{l} H_{0}: \mu=5 \\ H_{a}: \mu \neq 5 \end{array} $$ If the true conductivity is \(5.1,\) the liquid is not suitable for its intended use. You learn that the power of your test at the \(5 \%\) significance level against the alternative \(\mu=5.1\) is 0.23. (a) Explain in simple language what "power \(=0.23 "\) means in this setting. (b) You could get higher power against the same alternative with the same \(\alpha\) by changing the number of measurements you make. Should you make more measurements or fewer to increase power? (c) If you decide to use \(\alpha=0.10\) in place of \(\alpha=0.05\), with no other changes in the test, will the power increase or decrease? Justify your answer. (d) If you shift your interest to the alternative \(\mu=5.2\), with no other changes, will the power increase or decrease? Justify your answer.

A drug manufacturer claims that fewer than \(10 \%\) of patients who take its new drug for treating Alzheimer's disease will experience nausea. To test this claim, a significance test is carried out of $$ \begin{array}{l} H_{0}: p=0.10 \\ H_{a}: p<0.10 \end{array} $$ You learn that the power of this test at the \(5 \%\) significance level against the alternative \(p=0.08\) is 0.29 . (a) Explain in simple language what "power \(=0.29 "\) means in this setting. (b) You could get higher power against the same alternative with the same \(\alpha\) by changing the number of measurements you make. Should you make more measurements or fewer to increase power? Explain. (c) If you decide to use \(\alpha=0.01\) in place of \(\alpha=0.05\), with no other changes in the test, will the power increase or decrease? Justify your answer. (d) If you shift your interest to the alternative \(p=0.07\) with no other changes, will the power increase or decrease? Justify your answer.

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