/*! 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 Perform each of these steps. Ass... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Overweight Dogs A veterinary nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Six overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? $$ \begin{array}{l|llllll} \text { Before } & 42 & 53 & 48 & 65 & 40 & 52 \\ \hline \text { After } & 39 & 45 & 40 & 58 & 42 & 47 \end{array} $$

Short Answer

Expert verified
Yes, there is sufficient evidence that the dogs lost weight on the diet.

Step by step solution

01

State the Hypotheses and Identify the Claim

To determine if there's a significant weight loss, we start by setting up our null and alternative hypotheses. The null hypothesis \(H_0\) is that there is no difference in the weights before and after the diet. The alternative hypothesis \(H_1\) is that the weights have decreased after the diet.- Null hypothesis \(H_0\): \( \mu_{\text{after}} = \mu_{\text{before}} \)- Alternative hypothesis \(H_1\): \( \mu_{\text{after}} < \mu_{\text{before}} \)The claim is that the dogs lost weight, which aligns with the alternative hypothesis \(H_1\).
02

Find the Critical Value(s)

Given that we are using a significance level \( \alpha = 0.05 \) for a one-tailed test (since we are testing for 'less than'), we need the critical value from the t-distribution table for \(df = n-1 = 6-1 = 5\) degrees of freedom. Using the t-distribution table, the critical t-value for \(df = 5\) at a significance level of \(0.05\) is approximately \(-2.015\).
03

Compute the Test Value

Calculate the differences between the weights before and after for each dog. The differences are:\([-3, -8, -8, -7, 2, -5]\).* Calculate the mean difference \( \bar{d} \):\[\bar{d} = \frac{(-3) + (-8) + (-8) + (-7) + 2 + (-5)}{6} = \frac{-29}{6} = -4.833.\]* Calculate the standard deviation of the differences:Let \( s_d \) be the standard deviation of differences.* The formula for \( s_d \) is:\[s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}.\]After calculations, \( s_d \approx 3.766\).* Calculate the t-test value:\[t = \frac{\bar{d}}{(s_d / \sqrt{n})} = \frac{-4.833}{(3.766 / \sqrt{6})} \approx -3.219.\]
04

Make the Decision

Compare the test statistic computed \(t = -3.219\) with the critical value \(-2.015\). Since \(-3.219 < -2.015\), we reject the null hypothesis \(H_0\) at the 0.05 significance level.
05

Summarize the Results

There is sufficient statistical evidence at the \(0.05\) significance level to support the claim that the dogs lost weight after being subjected to the diet program.

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 t-Distribution
In hypothesis testing, particularly when dealing with small sample sizes, the t-distribution is critical. Unlike the normal distribution, the t-distribution accounts for the uncertainty inherent in estimating the population standard deviation from a sample. This increased uncertainty makes the tails of the t-distribution thicker than those of a normal distribution. This feature is especially useful when our sample size is small, like in the case of our dog diet example with just six observations.

You might ask why the t-distribution is used instead of the normal distribution. The reason lies in the sample size and variability. The smaller the sample, the less reliably you can estimate the standard deviation. Thus, the t-distribution offers a more cautious approach, adjusting for potential errors in this estimation.
  • It resembles the normal distribution, but is wider and fatter at the tails.
  • Approaches the normal distribution as sample size increases.
The knowledge of degrees of freedom (df), calculated as the number of cases minus one (n-1), is crucial for finding critical t-values, like in our dog's case with a df of 5.
What is the Null Hypothesis?
The null hypothesis, denoted as \(H_0\), is a fundamental concept in statistics. It represents the default assumption that there is no effect or difference. In hypothesis tests, our goal is to challenge this assumption and determine whether evidence is strong enough to reject it.

In the context of the dog diet study, the null hypothesis is that there is no difference in the average weights before and after the dogs were put on the diet. It is mathematically expressed as \( \mu_{\text{after}} = \mu_{\text{before}} \).

The null hypothesis acts as a baseline or starting point for statistical testing. It suggests that any observed differences in weights are due to random chance, not the diet. By discrediting the null hypothesis, we can argue in favor of the alternative hypothesis.
Exploring the Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\), is what researchers aim to support through their statistical tests. Unlike the null hypothesis, the alternative hypothesis suggests there is an effect or a difference that is not due to random chance.

In our dog diet scenario, the alternative hypothesis posits that the weights are less after the diet than before. Expressed as \( \mu_{\text{after}} < \mu_{\text{before}} \), it aligns with the claim that the dogs lost weight.

The alternative hypothesis is critical because it directly addresses the research question: "Did the diet result in weight loss for the dogs?" By using statistical evidence to reject the null hypothesis in favor of the alternative, researchers can substantiate their claims and draw meaningful conclusions.
Understanding Significance Level
The significance level, often denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. It is a threshold used to determine the critical value for hypothesis testing. A common choice for \( \alpha \) is 0.05, indicating a 5% risk of concluding that a difference exists when there is none.

Choosing the significance level is crucial, as it affects the stringency of the test. In the case of our dog diet study, a 0.05 significance level means that we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis, thus claiming the weight loss when none actually occurred.
  • Lower \( \alpha \) leads to stricter criteria for evidence.
  • Higher \( \alpha \) may increase the chance of Type I errors (false positives).
By using this significance level, we safeguard against making unjustified conclusions. Therefore, if we observe a test statistic beyond the critical value, we can confidently assert that the observed effect, such as weight loss, is statistically significant.

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

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A recent random survey of households found that 14 out of 50 householders had a cat and 21 out of 60 householders had a dog. At \(\alpha=0.05,\) test the claim that fewer household owners have cats than household owners who have dogs as pets.

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Summer reading programs are very popular with children. At the Citizens Library, Team Ramona read an average of 23.2 books with a standard deviation of 6.1. There were 21 members on this team. Team Beezus read an average of 26.1 books with a standard deviation of 2.3 . There were 23 members on this team. Did the variances of the two teams differ? Use \(\alpha=0.05\).

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Reducing Errors in Grammar A composition teacher wishes to see whether a new smartphone app will reduce the number of grammatical errors her students make when writing a two-page essay. She randomly selects six students, and the data are shown. At \(\alpha=0.025\), can it be concluded that the number of errors has been reduced? $$ \begin{array}{l|rrrrrr} \text { Student } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Errors before } & 12 & 9 & 0 & 5 & 4 & 3 \\ \hline \text { Errors after } & 9 & 6 & 1 & 3 & 2 & 3 \end{array} $$

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Pulse Rates of Identical Twins A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eight sets of twins were randomly selected. The rates are given in the table as number of beats per minute. At \(\alpha=0.01,\) is there a significant difference in the average pulse rates of twins? Use the \(P\) -value method. Find the \(99 \%\) confidence interval for the difference of the two. $$ \begin{array}{l|llllllll} \text { Twin A } & 87 & 92 & 78 & 83 & 88 & 90 & 84 & 93 \\ \hline \text { Twin B } & 83 & 95 & 79 & 83 & 86 & 93 & 80 & 86 \end{array} $$

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In a random sample of 200 men, 130 said they used seat belts. In a random sample of 300 women, 63 said they used seat belts. Test the claim that men are more safety-conscious than women, at \(\alpha=0.01\). Use the \(P\) -value method.

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.