/*! 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 3 Perform the appropriate hypothes... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 3

Perform the appropriate hypothesis test. A random sample of size \(n=13\) obtained from a population that is normally distributed results in a sample mean of 45.3 and sample standard deviation of \(12.4 .\) An independent sample of size \(n=18\) obtained from a population that is normally distributed results in a sample mean of 52.1 and sample standard deviation of 14.7 . Does this constitute sufficient evidence to conclude that the population means differ at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
No, there is not sufficient evidence to conclude that the population means differ at the \( \alpha = 0.05 \) level of significance.

Step by step solution

01

Formulate the Hypotheses

Determine the null and alternative hypotheses. The null hypothesis \( H_0 \) states that there is no difference in population means, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) states that the population means are different, i.e., \( \mu_1 \eq \mu_2 \).
02

Determine the Test Statistic

Since the population standard deviations are unknown and the samples are independent and normally distributed, use the t-test for comparing two means. The test statistic is calculated as: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}} \] where \( \bar{x}_1 = 45.3 \), \( \bar{x}_2 = 52.1 \), \( s_1 = 12.4 \), \( s_2 = 14.7 \), \( n_1 = 13 \), and \( n_2 = 18 \).
03

Calculate the Test Statistic

Plug in the values to calculate the test statistic: \[ t = \frac{(45.3 - 52.1)}{\sqrt{\frac{12.4^2}{13} + \frac{14.7^2}{18}}} \] Simplify the expression: \[ t = \frac{-6.8}{\sqrt{\frac{153.76}{13} + \frac{216.09}{18}}} = \frac{-6.8}{\sqrt{11.82 + 12.004}} = \frac{-6.8}{4.857} \approx -1.40 \]
04

Determine the Degrees of Freedom

The degrees of freedom for the test can be approximated using the formula: \[ df = \frac{\left(\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}\right)^2}{\frac{\left(\frac{s^2_1}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s^2_2}{n_2}\right)^2}{n_2-1}} \] Substituting the given values: \[ df = \frac{(\frac{153.76}{13} + \frac{216.09}{18})^2}{\frac{\left(\frac{153.76}{13}\right)^2}{12} + \frac{\left(\frac{216.09}{18}\right)^2}{17}} \approx 28.275 \] Use the nearest whole number, so \( df \approx 28 \).
05

Determine the Critical Value

For a two-tailed test at \( \alpha = 0.05 \) and \( df = 28 \), the critical value from the t-distribution table is approximately \( \pm 2.048 \).
06

Compare the Test Statistic and Critical Value

The test statistic calculated is \( t = -1.40 \) and the critical values are \( \pm 2.048 \). Since \( -2.048 < -1.40 < 2.048 \, \, \text{we fail to reject the null hypothesis}. \)
07

State the Conclusion

Since we fail to reject the null hypothesis, there is not sufficient evidence to conclude that the population means differ at the \( \alpha = 0.05 \) level of significance.

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.

t-test for two means
A t-test for two means helps determine if there is a significant difference between the means of two populations. This test is used when population standard deviations are unknown and the samples are independent and normally distributed. The test statistic is: } , {
statistical significance
Statistical significance tells us if our results are likely to have occurred by chance. In hypothesis testing, we often use a significance level (commonly denoted as \(\beta\)) to compare against our computed p-value. In this exercise, \(\beta = 0.05\), meaning we are 95% confident.
degrees of freedom
Degrees of freedom (df) dictate the shape of the t-distribution. They are calculated based on the sample size and are crucial for identifying critical values. In the exercise, df were calculated based on the size and variability of two populations, giving us df = 28.
critical value
The critical value is a threshold we use to compare our test statistic. It's determined by the significance level (a) and degrees of freedom (df). In this exercise, for % alpha = 0.05, the critical value for our two-tailed test with df = 28 is +-2.048.
population means comparison
To compare populations' means, we test if the difference observed from samples could be due to random error. In this problem, the hypothesis test showed that the sample means BO 45.3 and 52.1 were not sufficiently distinct to conclude population means differ at 伪 = 0.05 significance level.

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

Why do we use a pooled estimate of the population proportion when testing a hypothesis about two proportions? Why do we not use a pooled estimate of the population proportion when constructing a confidence interval for the difference of two proportions?

What is the typical age difference between husband and wife? The following data represent the ages of husbands and wives, based on results from the Current Population Survey. (a) What is the response variable in this study? (b) Is the sampling method dependent or independent? Explain. (c) Use the data to estimate the mean difference in age of husband and wives with \(95 \%\) confidence. Explain the technique that you used.

Walking in the Airport, Part II Do business travelers walk at a different pace than leisure travelers? Researcher Seth B. Young measured the walking speed of business and leisure travelers in San Francisco International Airport and Cleveland Hopkins International Airport. His findings are summarized in the table. $$ \begin{array}{lcc} \text { Type of Traveler } & \text { Business } & \text { Leisure } \\ \hline \text { Mean speed (feet per minute) } & 272 & 261 \\ \hline \begin{array}{l} \text { Standard deviation (feet per } \\ \text { minute) } \end{array} & 43 & 47 \\ \hline \text { Sample size } & 20 & 20 \\ \hline \end{array} $$ (a) Is this an observational study or a designed experiment? Why? (b) What must be true regarding the populations to use Welch's \(t\) -test to compare the means? (c) Assuming that the requirements listed in part (b) are satisfied, determine whether business travelers walk at a different speed from leisure travelers at the \(\alpha=0.05\) level of significance.

The Harris Poll conducted a survey in which they asked, 鈥淗ow many tattoos do you currently have on your body?鈥� Of the 1205 males surveyed, 181 responded that they had at least one tattoo. Of the 1097 females surveyed, 143 responded that they had at least one tattoo. Construct a 95% confidence interval to judge whether the proportion of males that have at least one tattoo differs significantly from the proportion of females that have at least one tattoo. Interpret the interval.

In Problems 9鈥�12, conduct each test at the a = 0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, (c) the critical value, and (d) the P-value. Assume that the samples were obtained independently using simple random sampling. Test whether \(p_{1}>p_{2}\). Sample data: \(x_{1}=368, n_{1}=541\), \(x_{2}=351, n_{2}=593\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Geometry

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.