Question

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

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

...

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 r

(e) interpret the decision in the context of the original claim.
at the 5% significance level, enough evidence to the claim tha
dogs is the mean annual cost of food for cats.

Explanation:

Step1: Recall the two - sample t - test for equal variances

The formula for the standardized test statistic (t - statistic) for two independent samples 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 \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\) is the pooled variance, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, \(n_1\) and \(n_2\) are the sample sizes, and \(\mu_1-\mu_2 = 0\) (under the null hypothesis that the population means are equal).

Step2: Calculate the pooled variance \(s_p^2\)

Given \(n_1 = 15\), \(s_1=31\), \(n_2 = 19\), \(s_2 = 28\)

$$
LATEXBLOCK0
$$

Step3: Calculate the standard error \(SE=\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}\)

$$
LATEXBLOCK1
$$

Step4: Calculate the t - statistic

\(\bar{x}_1 = 254\), \(\bar{x}_2=231\), \(\mu_1-\mu_2 = 0\)

$$
LATEXBLOCK2
$$

Answer:

The standardized test statistic \(t\) is \(\boldsymbol{2.27}\) (rounded to two decimal places).