H0: μ1=μ2
Ha: μ1<μ2
You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain a sample of size n1=11 with a mean of M1=90.9 and a standard deviation of SD1=19.8 from the first population. You obtain a sample of size n2=12 with a mean of M2=113.5 and a standard deviation of SD2=17.5 from the second population.
What is the critical value for this test? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to three decimal places.)
critical value = ___
What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic = ___
The test statistic is…
- in the critical region
- not in the critical region
This test statistic leads to a decision to…
- reject the null
- accept the null
- fail to reject the null
Answer:
Given:
Sample Mean 1 $(x\bar_1)=90.9$
Sample Standard Deviation 1 $(s_1)=19.8$
Sample Size 1 $(n_1)=11$
Sample Mean 2 $(x\bar_2)=113.5$
Sample Standard Deviation 2 $(s_2)=17.5$
Sample Size 2 $ (n_2)=12 $
Significance Level $ (\alpha) =0.10 $
Solution:
The null and alternative hypothesis:
$$H_0: \mu_1 = \mu_2 $$ $$ \quad H_a: \mu_1 < \mu_2$$
First, we need to find a degree of freedom to find out the critical value.
Therefore, the degree of freedom $(df):$
$$\therefore df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\left( \frac{s_1^2}{n_1} \right)^2 \div (n_1 – 1) + \left( \frac{s_2^2}{n_2} \right)^2 \div (n_2 – 1)} $$ $$ = \frac{\left( \frac{19.8^2}{11} + \frac{17.5^2}{12} \right)^2}{\left( \frac{19.8^2}{11} \right)^2 \div (11 – 1) + \left( \frac{17.5^2}{12} \right)^2 \div (12 – 1)} $$ $$ \approx 20.09 $
(1) The critical value $(t_c):$
$$\therefore t_c $$ $$ = t_{\frac{\alpha}{2}, \text{df}} $$ $$ = t_{\frac{0.10}{2}, 20.09} $$ $$ = -1.325 $$
(2) The test statistic $(t):$
$$ t = \frac{\overline{x}_1 – \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$ $$ = \frac{90.9 – 113.5}{\sqrt{\frac{19.8^2}{11} + \frac{17.5^2}{12}}} $$ $$ = -2.89 $$
(3) The test statistics is in the critical region. [ The test statistic is less than the critical value]
(4) The test statistic leads to a decision to reject the null [ The test statistic is less than the critical value]
Final Answer:
(1) The critical value $(t_c)=−1.325$
(2) The test statistic $(t)=−2.890$
(3) The test statistics is in the critical region.
(4) The test statistic leads to a decision to reject the null
