/*! 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 8 A simple random sample of size \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 8

A simple random sample of size \(n=320\) adults was asked their favorite ice cream flavor. Of the 320 individuals surveyed, 58 responded that they preferred mint chocolate chip. Do less than \(25 \%\) of adults prefer mint chocolate chip ice cream? Use the \(\alpha=0.01\) level of significance.

Short Answer

Expert verified
Less than 25% of adults prefer mint chocolate chip ice cream.

Step by step solution

01

- State the Hypotheses

Identify the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). \(H_0: p = 0.25\) (The proportion of adults who prefer mint chocolate chip ice cream is 25%) and \(H_a: p < 0.25\) (The proportion of adults who prefer mint chocolate chip ice cream is less than 25%).
02

- Calculate the Sample Proportion

Determine the sample proportion \(\hat{p}\) by dividing the number of adults who prefer mint chocolate chip by the total sample size: \(\hat{p} = \frac{58}{320} = 0.18125\).
03

- Calculate the Standard Error

Calculate the standard error (SE) of the sample proportion using the formula: \text{SE} = \sqrt{\frac{p(1-p)}{n}}\, where \(p = 0.25\) and \(n = 320\). \[ \text{SE} = \sqrt{\frac{0.25(1-0.25)}{320}} = 0.024206 \]
04

- Compute the Test Statistic

Calculate the test statistic (z) using the formula: \ z = \frac{\hat{p} - p}{\text{SE}} \: \[ z = \frac{0.18125 - 0.25}{0.024206} = -2.838 \]
05

- Determine the p-value

Find the p-value corresponding to the computed test statistic from the standard normal distribution table. For \(z = -2.838\), the p-value is approximately \(0.00226\).
06

- Compare the p-value with the Significance Level

Compare the p-value with the significance level \(\alpha = 0.01\). Since \(0.00226 < 0.01\), reject the null hypothesis.
07

- State the Conclusion

Conclude that there is sufficient evidence at the \(\alpha = 0.01\) level of significance to support the claim that less than 25% of adults prefer mint chocolate chip ice cream.

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 91影视!

Key Concepts

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

Null Hypothesis
In statistical hypothesis testing, the null hypothesis (\(H_0\)) represents the default assumption that nothing has changed or there is no effect. It's like saying the status quo holds true. In this exercise, the null hypothesis is: \(H_0: p = 0.25\). This means we assume the proportion of adults who prefer mint chocolate chip ice cream is 25%. Null hypotheses are crucial because they set a baseline which we test against.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) is what you want to test for. It's the challenger to the status quo. For this problem, the alternate hypothesis is: \(H_a: p < 0.25\). This means you are testing if the proportion of adults who prefer mint chocolate chip ice cream is less than 25%. The alternative hypothesis is directly linked to the research question and determines the type of test you will conduct (one-tailed or two-tailed).
P-value
A p-value helps determine the significance of the results in relation to the null hypothesis. It quantifies the probability of observing the test results, assuming the null hypothesis is true. A small p-value indicates that the observed data is inconsistent with the null hypothesis. For instance, in this ice cream problem, the calculated p-value was approximately 0.00226. If this p-value is less than the significance level (\(\alpha = 0.01\)), we reject the null hypothesis.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold set by the researcher which defines how much risk we are willing to take of rejecting a true null hypothesis. In simpler terms, it's your tolerance for making a Type I error (false positive). In this problem, the significance level is 0.01. This means we are willing to accept a 1% chance that we will incorrectly reject the null hypothesis. The p-value is compared to this significance level to make a decision.
Standard Error
The Standard Error (SE) reflects the variability or precision of the sample proportion as an estimate of the population proportion. For hypothesis testing, it's calculated using: \(\text{SE} = \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the assumed population proportion and \(n\) is the sample size. In this ice cream example, the SE is calculated as: \(\text{SE} = \sqrt{\frac{0.25(1-0.25)}{320}} = 0.024206\). Understanding SE helps in interpreting how much sample proportions might fluctuate.
Test Statistic
A test statistic is a standardized value derived from sample data used to test a hypothesis. For proportions, the test statistic for a z-test is calculated as: \(z = \frac{\hat{p} - p}{\text{SE}}\), where \(\hat{p}\) refers to the sample proportion, \(p\) to the hypothesized proportion, and \(SE\) to the standard error. In this exercise, the test statistic is: \(z = \frac{0.18125 - 0.25}{0.024206} = -2.838\). This value is then used to find the p-value and make an inference about the null hypothesis.

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

Decide whether the problem requires a confidence interval or hypothesis test, and determine the variable of interest. For any problem requiring a confidence interval, state whether the confidence interval will be for a population proportion or population mean. For any problem requiring a hypothesis test, write the null and alternative hypothesis. According to the Pew Research Center, \(55 \%\) of adult Americans support the death penalty for those convicted of murder. A social scientist wondered whether a higher proportion of adult Americans with at least a bachelor's degree support the death penalty for those convicted of murder.

In Problems \(15-22,(a)\) determine the null and alternative hypotheses, (b) explain what it would mean to make a Type I error, and (c) explain what it would mean to make a Type II error. Historically, the time to order and deliver a pizza at Jimbo's pizza was 48 minutes. Jim, the owner, implements a new system for ordering and delivering pizzas that he believes will reduce the time required to get a pizza to his customers.

Statistical Significance versus Practical Significance The manufacturer of a daily dietary supplement claims that its product will help people lose weight. The company obtains a random sample of 950 adult males aged 20 to 74 who take the supplement and finds their mean weight loss after 8 weeks to be 0.9 pound with standard deviation weight loss of 7.2 pounds. (a) State the null and alternative hypotheses. (b) Test the hypothesis at the \(\alpha=0.1\) level of significance. Is a mean weight loss of 0.9 pound significant? (c) Do you think that a mean weight loss of 0.9 pound is worth the expense and commitment of a daily dietary supplement? In other words, does the weight loss have any practical significance? (d) Test the hypothesis at the \(\alpha=0.1\) level of significance with \(n=40\) subjects. Assume the same sample statistics. Is a sample mean weight loss of 0.9 pound significantly more than 0 pound? What do you conclude about the impact of large samples on the hypothesis test?

Filling Bottles A certain brand of apple juice is supposed to have 64 ounces of juice. Because the penalty for underfilling bottles is severe, the target mean amount of juice is 64.05 ounces. However, the filling machine is not precise, and the exact amount of juice varies from bottle to bottle. The quality-control manager wishes to verify that the mean amount of juice in each bottle is 64.05 ounces so that she can be sure that the machine is not over- or underfilling. She randomly samples 22 bottles of juice, measures the content, and obtains the following data: $$ \begin{array}{llllll} \hline 64.05 & 64.05 & 64.03 & 63.97 & 63.95 & 64.02 \\ \hline 64.01 & 63.99 & 64.00 & 64.01 & 64.06 & 63.94 \\ \hline 63.98 & 64.05 & 63.95 & 64.01 & 64.08 & 64.01 \\ \hline 63.95 & 63.97 & 64.10 & 63.98 & & \\ \hline \end{array} $$ A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot does not show any outliers. (a) Should the assembly line be shut down so that the machine can be recalibrated? Use a 0.01 level of significance. (b) Explain why a level of significance of \(\alpha=0.01\) is more reasonable than \(\alpha=0.1 .\) [Hint: Consider the consequences of incorrectly rejecting the null hypothesis.

According to menstuff.org, \(22 \%\) of married men have "strayed" at least once during their married lives. (a) Describe how you might go about administering a survey to assess the accuracy of this statement. (b) A survey of 500 married men indicated that 122 have "strayed" at least once during their married life. Construct a \(95 \%\) confidence interval for the population proportion of married men who have strayed. Use this interval to assess the accuracy of the statement made by menstuff.org.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
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.