Question

The next 2 questions refer to the following scenario. Suppose the manager of a pet store wants to determine if there is a difference in the amount of money spent in the store, on average, by owners of dogs vs. owners of cats. Consider dog owners as group 1 and cat owners as group 2. Assume that the population standard deviations are equivalent between groups. Below are the sample data for the nine dog owners and the 9 cat owners. Test the hypothesis at alpha equals 1%.
The hypothesis at Ho: µ1 = µ2 vs. Ha: µ1 ≠ µ2.
Group1 Group 2
36 35.5
19 32.5
24.5 30
27 31.5
20 35.5
35 38
24.5 34.5
23.5 36
27.5 26

15) What is the p-value and test statistic used to test this hypothesis?
a) 0.0162 2.134
b) 0.9961 -2.134
c) 0.0339 3.044
d) 0.0756 3.044
e) 0.0077 -3.044

16) Is there a significant difference in the amount of money spent in the pet store, on average, by owners of dogs vs. c

Answer 1

Answer on Question #37888 – Math - Statistics and Probability

Assignment

The next 2 questions refer to the following scenario. Suppose the manager of a pet store wants to determine if there is a difference in the amount of money spent in the store, on average, by owners of dogs vs. owners of cats. Consider dog owners as group 1 and cat owners as group 2. Assume that the population standard deviations are equivalent between groups. Below are the sample data for the nine dog owners and the 9 cat owners. Test the hypothesis at alpha equals 1%.

The hypothesis at Ho: $\mu 1 = \mu 2$ vs. Ha: $\mu 1 \neq \mu 2$.

15) What is the p-value and test statistic used to test this hypothesis?

a) 0.0162 2.134

b) 0.9961 -2.134

c) 0.0339 3.044

d) 0.0756 3.044

e) 0.0077 -3.044

16) Is there a significant difference in the amount of money spent in the pet store, on average, by owners of dogs vs. c

Solution

Assumptions: 1. Both populations are normal. 2. The population standard deviations $\sigma_{1}$ and $\sigma_{2}$ are equal. From Excel data analysis (t-test: two sample assuming equal variances) we obtain p-value=0.0086 (two-tail), t-stat=-2.99, t critical two-tail=2.92. Since the p-value is 0.0086 is less than significance level 0.01, it can be concluded that there is a difference between means, so the evidence of $H_{a}$ is moderately strong.

Now we apply statistical methods. The hypothesis at Ho: $\mu 1 = \mu 2$ vs. Ha: $\mu 1 \neq \mu 2$. Let $X_{i}$ be observations from Group 1, $i = 1, \ldots, 9$, let $Y_{j}$ be observations from Group 2. Calculate $n_{1} = n_{2} = 9$, sample means $\bar{X} = 26.33$; $\bar{Y} = 33.28$, sample standard deviations $s_{1}^{2} = 5.91$, $s_{2}^{2} = 3.68$.

$$s _ {p o o l e d} ^ {2} = \frac {(n _ {1} - 1) s _ {1} ^ {2} + (n _ {2} - 1) s _ {2} ^ {2}}{n _ {1} + n _ {2} - 2} = \frac {8 * 5 . 9 1 + 8 * 3 . 6 8}{1 6} = 4. 7 9$$

We employ test statistic

$T = \frac{\bar{X} - \bar{Y}}{s_{\text{pooled}} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{(26.33 - 33.28)}{\sqrt{4.79} + \frac{\sqrt{2}}{3}} * 4.79 = -32.41$, d.f. $= n_1 + n_2 - 2 = 16$, we approximate the tabled value as $t_{0.01} = 2.92$, so the rejection region is $R$: $|T| \geq t$. It is true, this value lies in the rejection region $R$. Consequently, at the 0.01 level of significance, we reject the null hypothesis in favour of the alternative hypothesis that there is a significant difference in the amount of money spent in the pet store, on average, by owners of dogs vs. c