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 associations clai variances are equal. assume that the samples are random and independent, and the popul distributed. complete parts (a) through (e).

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

answer box(es) within your choice.
(round to two decimal places as needed.)
\\(\bigcirc\\) a. the rejection region is \\(t > \square\\).
\\(\bigcirc\\) b. the rejection region is \\(\square < t < \square\\).
\\(\bigcirc\\) c. the rejection region is \\(t < \square\\).
\\(\bigcirc\\) d. the rejection regions are \\(t < -2.04\\) and \\(t > 2.04\\).

(c) find the standardized test statistic.
\\(t = \square\\) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the standardized test statistic (t - statistic) for two - sample t - test with equal variances.

The formula for the t - statistic when testing the difference between two population means (\(\mu_1\) and \(\mu_2\)) 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 \(\mu_1-\mu_2 = 0\) (since the null hypothesis is that the means are equal, \(H_0:\mu_1=\mu_2\) or \(H_0:\mu_1-\mu_2 = 0\)).

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

Given \(\bar{x}_1 = 254\), \(s_1 = 31\), \(n_1 = 15\), \(\bar{x}_2=231\), \(s_2 = 28\), \(n_2 = 19\).
First, calculate \((n_1 - 1)s_1^2=(15 - 1)\times31^2=14\times961 = 13454\)
Then, calculate \((n_2 - 1)s_2^2=(19 - 1)\times28^2=18\times784 = 14112\)
The denominator for \(s_p^2\) is \(n_1 + n_2-2=15 + 19-2=32\)
So, \(s_p^2=\frac{13454 + 14112}{32}=\frac{27566}{32}\approx861.4375\)

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

\(\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{15}+\frac{1}{19}=\frac{19 + 15}{15\times19}=\frac{34}{285}\approx0.1193\)
\(SE=\sqrt{861.4375\times0.1193}=\sqrt{102.77}\approx10.137\)

Step4: Calculate the numerator \((\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)\)

Since \(\mu_1-\mu_2 = 0\) (null hypothesis), the numerator is \(\bar{x}_1-\bar{x}_2=254 - 231 = 23\)

Step5: Calculate the t - statistic

\(t=\frac{23}{10.137}\approx2.27\) (rounded to two decimal places)

Answer:

\(2.27\)