/*! 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 23 A recent study focused on the nu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A recent study focused on the number of times men and women who live alone buy takeout dinners in a month. The information is summarized below. $$\begin{array}{|lcc|}\hline \text { Statistic } & \text { Men } & \text { Women } \\\\\hline \text { Mean } & 24.51 & 22.69 \\\\\text { Standard deviation } & 4.48 & 3.86 \\\\\text { Sample size } & 35 & 40 \\\\\hline\end{array}$$ At the .01 significance level, is there a difference in the mean number of times men and women order takeout dinners in a month? What is the \(p\) -value?

Short Answer

Expert verified
No significant difference; \(p\)-value > 0.01, fail to reject \(H_0\).

Step by step solution

01

Define the Hypotheses

First, formulate the null and alternative hypotheses for the two-sample t-test. The null hypothesis (\(H_0\)) is that there is no difference in the means of takeout dinners ordered by men and women. Mathematically, \(H_0: \mu_{\text{Men}} = \mu_{\text{Women}}\).The alternative hypothesis (\(H_a\)) is that there is a difference, which implies \(H_a: \mu_{\text{Men}} eq \mu_{\text{Women}}\).
02

Determine the Significance Level

The significance level for this test is given as 0.01. This means we are using \(\alpha = 0.01\) to determine if we will reject the null hypothesis.
03

Calculate the Test Statistic

We will use the formula for the test statistic for two independent samples:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the given values: Mean for men \(\bar{x}_1 = 24.51\), standard deviation \(s_1 = 4.48\), sample size \(n_1 = 35\). Mean for women \(\bar{x}_2 = 22.69\), standard deviation \(s_2 = 3.86\), sample size \(n_2 = 40\).Thus,\[ t = \frac{24.51 - 22.69}{\sqrt{\frac{4.48^2}{35} + \frac{3.86^2}{40}}} \approx \frac{1.82}{\sqrt{0.5727 + 0.3726}} \approx \frac{1.82}{0.9783} \approx 1.86 \]
04

Find the Critical Value and P-Value

For a two-tailed test at \(\alpha = 0.01\) and degrees of freedom approximated using the formula for unequal variances, the critical value from the t-distribution table should be determined. However, generally beyond simpler calculations, software is used to obtain precise p-values. Here, approximate by stating if \(|t|\) exceeds the critical value around +/- 2.637 (for >30 df). Calculating the p-value manually requires numerical/software help suggesting the t-value's probability is found near about 0.067. Thus, ensure referring to p-value tables/softwares.
05

Decision on Hypotheses

Since the computed t-value 1.86 is less than the critical t-value of approximately 2.637, and the p-value > 0.01, we do not have sufficient evidence to reject the null hypothesis. There is no significant difference in the mean number of takeout dinners ordered by men and women at the 0.01 significance level.

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.

Understanding the Two-Sample T-Test
In the context of hypothesis testing, a two-sample t-test is a statistical method used to compare the means from two independent groups. It's particularly useful when you want to determine if there's a significant difference between the means of these two groups. Here, we're examining whether men and women, who live alone, differ in their monthly takeout dinner orders.

This test involves a few critical assumptions:
  • Both samples are random and independent.
  • The data in each group is approximately normally distributed.
  • The variances from both populations are equal, although there are adjustments that can be made when this isn't the case.
Each of these conditions helps ensure the validity of the test, leading to more reliable results.
Calculating the P-Value
The p-value is crucial in determining if your hypothesis test results are statistically significant. In simpler terms, it tells us the probability of obtaining test results at least as extreme as the ones you observed, assuming that the null hypothesis is true.

For the two-sample t-test, after calculating the test statistic (which in this scenario is approximately 1.86), we compare it against the distribution under the null hypothesis. The p-value thus derived helps inform our decision-making process when contemplating the null. Typically, a smaller p-value indicates stronger evidence against the null hypothesis. Here, a p-value near 0.067 suggests not rejecting the null hypothesis, as it is greater than the chosen significance level of 0.01.
Significance Level
In hypothesis testing, the significance level, usually denoted by \( \alpha \), represents the threshold for determining whether a result is statistically significant. Often set at 0.05, or 5%, it indicates the probability of rejecting the null hypothesis when it is, in fact, true. Regarding critical decision-making, this is referred to as the Type I error rate.

In this exercise, we use a significance level of 0.01, meaning we're opting for a more stringent criterion. It implies that we are willing to accept a 1% chance of rejecting the null hypothesis incorrectly. Using such a low \( \alpha \) level requires particularly strong evidence against the null hypothesis for it to be rejected, reflecting a more conservative approach to data analysis.
Formulating Null and Alternative Hypotheses
In any hypothesis test, setting clear null and alternative hypotheses is one of the first crucial steps. The null hypothesis, often denoted as \( H_0 \), typically suggests no effect or no difference in the context of the two sampled groups. It's the baseline statement that we attempt to refute with our statistical test.

Meanwhile, the alternative hypothesis, \( H_a \), is what you might call the research hypothesis. It suggests a deviation from the null hypothesis, indicating the presence of an effect or a difference. In our case, the null hypothesis states that the mean takeout orders by men and women are the same, while the alternative hypothesis asserts there is a difference in means. Clear formulation of these hypotheses guides the entire analytic process and influences the interpretation of results.

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

Lester Hollar is Vice President for Human 91Ó°ÊÓ for a large manufacturing company. In recent years he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the \(p\) -value.

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \pi_{1}=\pi_{2} \\\H_{1}: \pi_{1} \neq \pi_{2}\end{array}$$ A sample of 200 observations from the first population indicated that \(X_{1}\) is \(170 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(110 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

The owner of Bun 'N' Run Hamburger wishes to compare the sales per day at two locations. The mean number sold for 10 randomly selected days at the Northside site was 83.55, and the standard deviation was \(10.50 .\) For a random sample of 12 days at the Southside location, the mean number sold was 78.80 and the standard deviation was \(14.25 .\) At the .05 significance level, is there a difference in the mean number of hamburgers sold at the two locations? What is the \(p\) -value?

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \mu_{1}=\mu_{2} \\\H_{1}: \mu_{1} \neq \mu_{2}\end{array}$$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of \(5 .\) At the .05 significance level, is there a difference between the population means?

As part of a study of corporate employees, the Director of Human 91Ó°ÊÓ for PNC, Inc. wants to compare the distance traveled to work by employees at their office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month, with a standard deviation of 30 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month, with a standard deviation of 26 miles per month. At the .05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees? Use the five-step hypothesis- testing procedure.

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.