/*! 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 51 Determine whether it is appropri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether it is appropriate to use the normal distribution to estimate the p-value. If it is appropriate, use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from a random sample and use a \(5 \%\) significance level. Test \(H_{0}: p=0.2\) vs \(H_{a}: p \neq 0.2\) using the sample results \(\hat{p}=0.26\) with \(n=1000\)

Short Answer

Expert verified
Using a normal distribution is appropriate. The test statistic is approximately 4.74, the P-value is approximately 0, and we reject the null hypothesis. Therefore, we conclude that there's significant evidence to support the claim that the population proportion is not 0.2.

Step by step solution

01

Calculate Test Statistic

To perform the hypothesis test, we first need to calculate the test statistic. This is done by subtracting the hypothesized population proportion from the sample proportion, and dividing by the standard deviation of the sampling distribution of the sample proportion. The standard deviation \(\sigma\) can be found as \(\sigma = \sqrt{p(1-p)/n}\), where \(p\) is the hypothesized population proportion and \(n\) is the sample size. Here, \(\sigma = \sqrt{0.2(1-0.2)/1000} = 0.012649\). The test statistic \(z\) is then given by \(z = (\hat{p} - p) / \sigma\), where \(\hat{p}\) is the sample proportion. Here, \(z = (0.26 - 0.20) / 0.012649 = 4.74.
02

Determine the P-value

The P-value is the probability of obtaining a result as extreme as our sample result, if the null hypothesis is true. As our alternative hypothesis is \(p \neq 0.2\), this is a two-tailed test, so we want the probability that the test statistic is less than -4.74 or greater than 4.74. From the standard normal table, the probability of getting a z-score less than -4.74 or greater than 4.74 is essentially zero, so the P-value is approximately 0.
03

Interpret the Result

We compare our P-value to our significance level of 0.05. As our P-value is smaller than 0.05, we reject the null hypothesis. This means we have significant evidence to suggest that the population proportion is not 0.2.

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.

Normal Distribution
The normal distribution is a continuous probability distribution that is symmetrical on both sides of its mean, resembling the shape of a bell when graphed. This is why it's also known as the 'bell curve.' The characteristics of a normal distribution include that about 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three.

To utilize the normal distribution for hypothesis testing, the sampling distribution (distribution of sample statistics like mean or proportion) should be approximately normally distributed. This condition is frequently met when the sample size is large due to the Central Limit Theorem, which states that with a sufficiently large sample size, the distribution of the sample means will be approximately normal, regardless of the population's distribution shape.

In the context of the exercise, because we have a large sample size (n=1000), it is appropriate to assume that the sampling distribution of the sample proportion is normally distributed. This allows us to use the normal distribution to estimate the p-value and perform the hypothesis test. The exercise improvements would suggest explaining this background to ensure students understand when and why the normal distribution is applied in hypothesis testing.
P-Value
The p-value in hypothesis testing is a probability that measures the strength of the evidence against the null hypothesis. It is defined as the probability of obtaining test results at least as extreme as the observed results, assuming that the null hypothesis is correct.

A low p-value indicates that the observed data are unlikely under the null hypothesis and provides evidence to reject the null hypothesis in favor of the alternative hypothesis. Conversely, a high p-value suggests that the data are consistent with the null hypothesis.

In practice, if the p-value is less than or equal to the significance level (commonly \(\alpha = 0.05\)), we reject the null hypothesis. In the exercise provided, the p-value is approximately 0, which is much lower than the significance level of 0.05, indicating strong evidence against the null hypothesis and leading us to reject it. Clear communication about what a p-value represents and how it is applied in decision-making is crucial for student comprehension, as reinforced by the exercise improvement tips.
Test Statistic
A test statistic is a value calculated from the sample data that is used in a hypothesis test. It is the tool by which we decide whether to accept or reject the null hypothesis. The test statistic is used to measure the extremeness of the data given the assumption that the null hypothesis is true.

The test statistic is calculated based on the sample data and the hypothesized parameter under the null hypothesis. For tests about a population proportion, the z-test statistic is commonly used, which can be calculated when the sample size is large enough to assume that the sampling distribution of the proportion is normally distributed.

In the provided exercise, the test statistic \(z\) helps determine whether the observed sample proportion \(\hat{p}\) is significantly different from the hypothesized population proportion. It forms the basis for calculating the p-value and making decisions regarding the null hypothesis. Such explanations align with exercise improvement advice which promotes deep understanding of why and how a test statistic is used in hypothesis testing.
Population Proportion
Population proportion refers to the percentage or fraction of individuals in the population that exhibit a certain characteristic or attribute. It is denoted by \(p\) and is a parameter that we often wish to estimate or test hypotheses about using sample data.

In hypothesis testing, we often have a null hypothesis that assumes a specific value for the population proportion. The sample proportion \(\hat{p}\), calculated from our sample data, serves as an estimate for this population proportion and is used to determine the sample's deviation from the hypothesized proportion.

The exercise provided uses the population proportion in the null hypothesis (\(H_{0}: p=0.2\)) and compares it against the alternative hypothesis that the proportion is different from 0.2 based on the sample results. Understanding the role of population proportion in hypothesis tests allows students to appreciate its importance in the field of inferential statistics and adds depth to the discussion, in line with the advice provided for exercise enhancement.

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

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=5.2, s_{1}=2.7, n_{1}=10\) and \(\bar{x}_{2}=4.9, s_{2}=2.8, n_{2}=8 .\)

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

Metal Tags on Penguins and Breeding Success Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. Exercise 6.148 investigates the survival rate of the penguins. The scientists also studied the breeding success of the metal- and electronic-tagged penguins. Metal-tagged penguins successfully produced offspring in \(32 \%\) of the 122 total breeding seasons, while the electronic-tagged penguins succeeded in \(44 \%\) of the 160 total breeding seasons. Construct a \(95 \%\) confidence interval for the difference in proportion successfully producing offspring \(\left(p_{M}-p_{E}\right)\). Interpret the result.

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.