Question

For Exercises 2 through \(12,\) 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? $$ \begin{array}{l|cccccc}{\text { Before }} & {42} & {53} & {48} & {65} & {40} & {52} \\ \hline \text { After } & {39} & {45} & {40} & {58} & {42} & {47}\end{array} $$

Short answer

The data supports the claim that the dogs lost weight on the diet.

Step-by-step solution

State the Hypotheses

Begin by stating the null and alternative hypotheses. For this problem, the null hypothesis ( H_0 ) is that there is no difference in the mean weight before and after the diet, which means the diet has no effect. The alternative hypothesis ( H_1 ) is that there is a difference, suggesting that the diet leads to weight loss. Formally: Null Hypothesis ( H_0 ): μ_d = 0 Alternative Hypothesis ( H_1 ): μ_d > 0 (since we expect weight loss, we test if the mean difference is greater than 0).

Identify the Claim

The claim is the alternative hypothesis that the diet helps in weight reduction, expressed as μ_d > 0 .

Find the Critical Values

With a significance level of α = 0.05 , determine the critical value for a one-tailed t-test with 5 degrees of freedom (since N = 6 - 1 ). Using a t-table, the critical value ( cv ) for α = 0.05 and 5 degrees of freedom is approximately 2.015 .

Compute the Test Value

First, calculate the differences in weights before and after the diet for each dog. The differences are:(42-39), (53-45), (48-40), (65-58), (40-42), (52-47),giving: 3, 8, 8, 7, -2, 5.Calculate the mean of these differences (ar{d}) and the standard deviation (s_d).Mean difference (ar{d}): (3 + 8 + 8 + 7 - 2 + 5) / 6 = 4.833Standard deviation (s_d) using standard deviation formula results in 3.768 approximately.Now, compute the test statistic (t) using:\[ t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{4.833}{3.768/\sqrt{6}} \approx 4.204 \]

Make the Decision

Compare the computed test value 4.204 with the critical value 2.015 . Since 4.204 is greater than 2.015 , we reject the null hypothesis.

Summarize the Results

Since the null hypothesis is rejected, there is sufficient evidence at the 0.05 level of significance to support the claim that the diet program causes weight loss in dogs.

Key concepts

Understanding the Null Hypothesis

In any hypothesis testing, the null hypothesis is like our starting point. We assume that there is no effect or change caused by the experiment. For example, in our dog weight loss study, the null hypothesis \((H_0)\) states that the new diet does not cause weight loss, meaning the average weight before and after dieting is the same. Mathematically, for difference in means, it is expressed as \( \mu_d = 0 \). This means that the average weight difference of the dogs before and after the diet is zero, indicating no weight loss. We use the null hypothesis as our baseline, and attempt to find evidence that can 'disprove' it. That's the goal of our test: to see if there's enough evidence against \( H_0 \) to consider an alternative.

Exploring the Alternative Hypothesis

The alternative hypothesis \((H_1)\) is everything the null hypothesis is not. It represents the new claim or effect we are testing for. In the dieting dogs example, our alternative hypothesis suggests that the diet does work, causing weight loss. So, \( \mu_d > 0 \) asserts that the average weight difference \( \bar{d} \) is positive.Choosing the direction of the alternative hypothesis matters. In this example, a one-tailed (or directional) alternative hypothesis is used because we are testing for a specific effect: weight loss. This contrasts with a two-tailed hypothesis, which might be used if we just wanted to see any change, whether it was gain or loss in weight.

Understanding the T-Test

A t-test is a statistical tool used to determine if there is a significant difference between the means of two groups. In our exercise with the dogs, we're using a paired sample t-test. This is specific to cases where we are comparing two related samples, like the weights before and after the diet.To perform the t-test, we calculate the mean difference \( \bar{d} \) of weights before and after the diet and the standard deviation \( s_d \) of these differences. Using these, the t-test statistic is calculated with the formula:\[ t = \frac{\bar{d}}{s_d/\sqrt{n}} \]where \( n \) is the number of dogs in this case. The computed test statistic helps us understand how far our sample's observations diverge from the null hypothesis.

Deciphering Critical Values

Critical values are thresholds we use to decide whether to reject the null hypothesis. In our one-tailed test with a significance level \( \alpha = 0.05 \), we find the critical value for \( n-1 \) degrees of freedom using t-distribution tables. For our dog experiment with 6 dogs, we have \( 5 \) degrees of freedom \((n-1=5)\). In this case, the critical value is approximately \( 2.015 \). This value is crucial because:- If our calculated t-value exceeds this number, we reject the null hypothesis.- If it does not, we fail to reject the null hypothesis. This provides a clear decision-making framework for evaluating the results of hypothesis testing.

Significance Level Explained

The significance level \( \alpha \) is a pre-set threshold used to decide how confident we are in rejecting the null hypothesis. Typically, values like 0.05 or 0.01 are used, meaning we accept a 5% or 1% chance of being wrong, respectively.In our dog weight study, we set \( \alpha = 0.05 \), which implies a 5% risk of concluding that the diet works when it actually doesn't (Type I error). At this level, if the p-value (probability the null hypothesis is true) is less than or equal to \( \alpha \), the results are considered significant. This significance indicates strong enough evidence to support the alternative hypothesis.