/*! 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 5 A horse trainer teaches horses t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 5

A horse trainer teaches horses to jump by using two methods of instruction. Horses being taught by method A have a lead horse that accompanies each jump. Horses being taught by method B have no lead horse. The table shows the number of training sessions required before each horse performed the jumps properly. $$ \begin{array}{l|llllllllllll} \hline \text { Method A } & 28 & 35 & 19 & 41 & 37 & 31 & 38 & 40 & 25 & 27 & 36 & 43 \\ \hline \text { Method B } & 42 & 33 & 26 & 24 & 44 & 46 & 34 & 20 & 48 & 39 & 45 & \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the claim that there is no difference between the training session distributions.

Short Answer

Expert verified
Conduct a two-sample t-test; compare t-statistic to critical value. If absolute t exceeds critical, reject \(H_0\). Results show whether distributions differ.

Step by step solution

01

Formulate the Hypotheses

We need to formulate the null and alternative hypotheses for this test. The null hypothesis (\(H_0\)) states that there is no difference in the distribution of the number of training sessions required between methods A and B. The alternative hypothesis (\(H_1\)) states that there is a difference. Mathematically, \(H_0: \mu_A = \mu_B\) and \(H_1: \mu_A eq \mu_B\).
02

Choose the Test and Level of Significance

Since we are comparing the means of two independent samples with unknown variances, we will use a two-sample t-test. The level of significance is given as 5\%, or \(\alpha = 0.05\).
03

Calculate the Sample Means

Calculate the mean for each group: Method A and Method B. \[\bar{x}_A = \frac{28 + 35 + 19 + 41 + 37 + 31 + 38 + 40 + 25 + 27 + 36 + 43}{12},\]\[\bar{x}_B = \frac{42 + 33 + 26 + 24 + 44 + 46 + 34 + 20 + 48 + 39 + 45}{11}.\]
04

Calculate Sample Standard Deviations

Calculate the standard deviations for each group. This will involve finding the variance for Method A and Method B, then taking the square root. For Method A, this is: \[s_A = \sqrt{\frac{1}{n-1}\sum{(x_i - \bar{x}_A)^2}},\]and similarly for Method B: \[s_B = \sqrt{\frac{1}{n-1}\sum{(x_i - \bar{x}_B)^2}}.\]
05

Calculate the Test Statistic

The formula for the t-statistic in a two-sample t-test is: \[t = \frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}},\]where \(n_A\) and \(n_B\) are the sample sizes.
06

Determine the Degrees of Freedom

Degrees of freedom for a two-sample t-test can be calculated using: \[df = \frac{\left(\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}\right)^2}{\frac{\left(\frac{s_A^2}{n_A}\right)^2}{n_A-1} + \frac{\left(\frac{s_B^2}{n_B}\right)^2}{n_B-1}}.\]
07

Decision Rule and Conclusion

Use the t-distribution table to find the critical t-value for a two-tailed test at \(\alpha = 0.05\) and the calculated degrees of freedom. Compare the calculated t-statistic to this critical value: if \(|t|\) is greater than the critical value, reject \(H_0\).

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.

Hypothesis Testing
Hypothesis testing is a fundamental process in statistics used to make inferences about population parameters using sample data. It starts with forming two statements: a null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_1\)). The null hypothesis usually posits that there is no effect or difference, while the alternative suggests there is an effect or difference. In our example, the null hypothesis states there are no differences in the number of training sessions needed between two teaching methods for horses.
  • \(H_0: \mu_A = \mu_B\) - The means for both methods are equal.
  • \(H_1: \mu_A eq \mu_B\) - The means for both methods are not equal.
Once these hypotheses are established, statistical tests, such as a t-test in this case, help determine which hypothesis is more likely to be true given the data.
Level of Significance
The level of significance, often denoted as \(\alpha\), is a crucial component in hypothesis testing. It indicates the probability of rejecting the null hypothesis when it is true, essentially quantifying the risk of a Type I error. Commonly used significance levels are 0.05, 0.01, and 0.10. In this exercise, a 5% level of significance (\(\alpha = 0.05\)) is used, meaning there's a 5% chance of incorrectly concluding that there is a difference between the training methods' means.
This critical value helps determine the threshold or cutoff in hypothesis testing. If the calculated test statistic exceeds this threshold, the null hypothesis is rejected.
Sample Mean
The sample mean is an essential statistical measure representing the average value in a dataset and provides a central tendency measure. It is calculated by summing all the observed values and dividing by the number of observations. In the context of our exercise, the sample means for Method A and Method B offer insights into their average number of training sessions.
The formulas for sample means are: \[\bar{x}_A = \frac{28 + 35 + 19 + 41 + 37 + 31 + 38 + 40 + 25 + 27 + 36 + 43}{12}\] \[\bar{x}_B = \frac{42 + 33 + 26 + 24 + 44 + 46 + 34 + 20 + 48 + 39 + 45}{11}\]
Calculating these means provides an initial comparison point for the two teaching methodologies before applying further statistical analysis.
Standard Deviation
Standard deviation is a measure of data spread relative to the mean, reflecting data variability. It indicates how much individual data points deviate from the average. A higher standard deviation means more variability, while a lower one indicates data points are closer to the mean.
In our test, computing the standard deviation for training sessions using methods A and B was necessary to determine how much the number of sessions varied among the horses within each method. It involves several steps:
  • Calculate the variance, which represents the average of squared deviations from the mean.
  • The standard deviation is the square root of the variance, giving a sense of spread in the same units as the data.
By calculating the standard deviation, we can further understand the distribution differences which contribute to determining the validity of the initial 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

A psychologist has developed a mental alertness test. She wishes to study the effects (if any) of type of food consumed on mental alertness. Twenty-one volunteers were randomly divided into two groups. Both groups were told to eat the amount they usually eat for lunch at noon. At 2:00 P.M., all subjects were given the alertness test. Group A had a low-fat lunch with no red meat, lots of vegetables, carbohydrates, and fiber. Group B had a high-fat lunch with red meat, vegetable oils, and low fiber. The only drink for both groups was water. The test scores are shown below. $$ \begin{array}{l|cccccccccccc} \hline \text { Group A } & 76 & 93 & 52 & 81 & 68 & 79 & 88 & 90 & 67 & 85 & 60 & \\ \hline \text { Group B } & 44 & 57 & 60 & 91 & 62 & 86 & 82 & 65 & 96 & 42 & 68 & 98 \\ \hline \end{array} $$ Use a \(1 \%\) level of significance to test the claim that there is no difference in mental alertness distributions based on type of lunch.

The following data represent annual percentage returns on Vanguard Total Bond Index for a sequence of recent years. This fund represents nearly all publicly traded U.S. bonds (Reference: Morningstar Mutual Fund Analysis). $$ \begin{array}{llllllllllll} 7.1 & 9.7 & -2.7 & 18.2 & 3.6 & 9.4 & 8.6 & -0.8 & 11.4 & 8.4 & 8.3 & 0.8 \end{array} $$ (i) Convert this sequence of numbers to a sequence of symbols \(\mathrm{A}\) and \(\mathrm{B}\), where A indicates a value above the median and B a value below the median. (ii) Test the sequence for randomness about the median. Use \(\alpha=0.05\).

An army psychologist gave a random sample of seven soldiers a test to measure sense of humor and another test to measure aggressiveness. Higher scores mean greater sense of humor or more aggressiveness. $$ \begin{array}{l|rrrrrrr} \hline \text { Soldier } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Score on humor test } & 60 & 85 & 78 & 90 & 93 & 45 & 51 \\ \text { Score on aggressiveness test } & 78 & 42 & 68 & 53 & 62 & 50 & 76 \\ \hline \end{array} $$ (i) Ranking the data with rank 1 for highest score on a test, make a table of ranks to be used in a Spearman rank correlation test. (ii) Using a \(0.05\) level of significance, test the claim that rank in humor has a monotone-decreasing relation to rank in aggressiveness.

A cognitive aptitude test consists of putting together a puzzle. Eleven people in group A took the test in a competitive setting (first and second to finish received a prize). Twelve people in group B took the test in a noncompetitive setting. The results follow (in minutes required to complete the puzzle). $$ \begin{array}{l|cccccccccccc} \hline \text { Group A } & 7 & 12 & 10 & 15 & 22 & 17 & 18 & 13 & 8 & 16 & 11 & \\ \hline \text { Group B } & 9 & 16 & 30 & 11 & 33 & 28 & 19 & 14 & 24 & 27 & 31 & 29 \\ \hline \end{array} $$ Use a \(5 \%\) level of significance to test the claim that there is no difference in the distributions of time to complete the test.

Sand and clay studies were conducted at the West Side Field Station of the University of California (Reference: Professor D. R. Nielsen, University of California, Davis). Twelve consecutive depths, each about \(15 \mathrm{~cm}\) deep, were studied and the following percentages of clay in the soil were recorded. $$ \begin{array}{llllllllllllll} 47.4 & 43.4 & 48.4 & 42.6 & 41.4 & 40.7 & 46.4 & 44.8 & 36.5 & 35.7 & 33.7 & 42.6 \end{array} $$ (i) Convert this sequence of numbers to a sequence of symbols \(\mathrm{A}\) and \(\mathrm{B}\), where A indicates a value above the median and \(\mathrm{B}\) a value below the median. (ii) Test the sequence for randomness about the median. Use \(\alpha=0.05\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

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.