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 population distributed. complete parts (a) through (e).
dogs:
$\bar{x}_1 = \\$254$
$s_1 = \\$31$
$n_1 = 15$
cats:
$\bar{x}_2 = \\$231$
$s_2 = \\$28$
$n_2 = 19$
a. the rejection region is $t > \square$.
b. the rejection region is $\square < t < \square$.
c. the rejection region is $t < \square$.
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.
$\square$ the null hypothesis because the test statistic $\square$ in a/the rejection regi

Explanation:

Step1: Identify Hypothesis Test Type

This is a two - sample t - test for means with equal variances. The null hypothesis \(H_0:\mu_1=\mu_2\) and the alternative hypothesis \(H_a:\mu_1
eq\mu_2\) (since we are testing if the means are the same, a two - tailed test).

Step2: Determine Rejection Region

For a two - tailed test with \(\alpha = 0.05\), the degrees of freedom \(df=n_1 + n_2-2=15 + 19-2 = 32\). Looking up the t - critical value in the t - distribution table, \(t_{\alpha/2,df}=t_{0.025,32}\approx2.04\). So the rejection regions are \(t < - 2.04\) and \(t>2.04\).

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 the 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, calculate the t - statistic:

$$t=\frac{(254 - 231)-0}{\sqrt{861.4375\times(\frac{1}{15}+\frac{1}{19})}}=\frac{23}{\sqrt{861.4375\times(\frac{19 + 15}{15\times19})}}=\frac{23}{\sqrt{861.4375\times\frac{34}{285}}}=\frac{23}{\sqrt{\frac{861.4375\times34}{285}}}=\frac{23}{\sqrt{\frac{29288.875}{285}}}=\frac{23}{\sqrt{102.76798}}\approx\frac{23}{10.137}\approx2.27$$

Step4: Decision Rule

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

Answer:

(d) Reject the null hypothesis because the test statistic \(2.27\) is in the rejection region (\(t>2.04\)). So the first dropdown is "Reject", the second dropdown is "is".