Question

a pet association claims that the mean annual costs of food for dogs and cats are the same. the two types of pets are shown below. at α = 0.05, can you reject the pet association’s claim? variances are equal. assume that the samples are random and independent, and the populations are distributed. complete parts (a) through (e).

dogs	cats
\\(s_1 = \\$31\\)	\\(s_2 = \\$28\\)
\\(n_1 = 15\\)	\\(n_2 = 19\\)

the rejection region is t > blank.
d. the rejection regions are t < -2.04 and t > 2.04.
(c) find the standardized test statistic.
t = 2.27 (round to two decimal places as needed.)
(d) decide whether to reject or fail to reject the null hypothesis.
reject the null hypothesis because the test statistic is in a/the rejection region.
(e) interpret the decision in the context of the original claim.
at the 5% significance level, dropdown enough evidence to dropdown the claim that the mean annual cost of food for dogs is dropdown the mean annual cost of food for cats.

Explanation:

Step1: Recall Hypothesis Test Context

We have a two - sample t - test (since population variances are equal, samples are random, independent, and populations are normally distributed) to test the claim that the mean annual costs of food for dogs and cats are the same. The null hypothesis \(H_0:\mu_1=\mu_2\) and the alternative hypothesis \(H_a:\mu_1
eq\mu_2\) (two - tailed test) with \(\alpha = 0.05\).

Step2: Determine Degrees of Freedom

The degrees of freedom for a two - sample t - test with equal variances is \(df=n_1 + n_2-2=15 + 19-2 = 32\). For a two - tailed test with \(\alpha=0.05\) and \(df = 32\), the critical values (from t - distribution table or calculator) are \(t_{\alpha/2,df}=t_{0.025,32}\approx\pm2.04\). So the rejection regions are \(t < - 2.04\) or \(t>2.04\).

Step3: Calculate Test Statistic

The formula for the standardized test statistic (t - statistic) for two - sample t - test with equal variances is:

$$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}}$$

where \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\), and \(\mu_1-\mu_2 = 0\) (under null hypothesis).

First, calculate \(s_p^2\):

$$s_p^2=\frac{(15 - 1)\times31^2+(19 - 1)\times28^2}{15 + 19-2}=\frac{14\times961+18\times784}{32}=\frac{13454+14112}{32}=\frac{27566}{32}\approx861.4375$$

Then, \(\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{861.4375\times(\frac{1}{15}+\frac{1}{19})}=\sqrt{861.4375\times(\frac{19 + 15}{15\times19})}=\sqrt{861.4375\times\frac{34}{285}}=\sqrt{861.4375\times0.1193}\approx\sqrt{102.7}\approx10.13\)
\(\bar{x}_1-\bar{x}_2=254 - 231 = 23\)

$$t=\frac{23-0}{10.13}\approx2.27$$

Step4: Decision Rule

We compare the test statistic \(t = 2.27\) with the critical values. Since \(2.27>2.04\), the test statistic lies in the rejection region (\(t>2.04\)).

Step5: Interpret the Decision

Since the test statistic is in the rejection region, we reject the null hypothesis. In the context of the original claim, at the 5% significance level, there is enough evidence to reject the claim that the mean annual cost of food for dogs is equal to the mean annual cost of food for cats.

Answer:

For part (e): At the 5% significance level, there is enough evidence to reject the claim that the mean annual cost of food for dogs is equal to the mean annual cost of food for cats. (Filling the blanks: "there is", "reject", "equal to")