Question

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

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

Step-by-step solution

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 \(H_1\).

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 \(-2.015\).

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:\[t = \frac{\bar{d}}{(s_d / \sqrt{n})} = \frac{-4.833}{(3.766 / \sqrt{6})} \approx -3.219.\]

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.

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.

Key concepts

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.