Question

Assume that you have a sample of n = 6, with the sample mean X, = 46, and a sample standard deviation of S7 = 4, and you have an independent sample of n2 = 7 from another population with a sample mean of X2 = 35 and the sample standard deviation S2 = 5. Assuming the population variances are equal, at the 0.01 level of significance, is there evidence that Hy >H2? Determine the hypotheses. Choose the correct answer below. O A. Ho: H1 H2 H:H1 H2 B. Ho H1 H2 H:H7 > H2 OD. Ho: H1 H2 H:H1 = 12 C. Ho: M1 > H2 H:H1 H2 Find the test statistic. tstat = 4.32 (Round to two decimal places as needed.) Find the p-value. p-value = (Round to three decimal places as needed.)

Answer 1

$\\\bar{X_1} = 46 \\\bar{X_2} = 35 \\S_1 = 4 \\S_2 = 5 \\S_1^2 = 16.0 \\S_2^2 = 25.0 \\n_1 = 6 \\n_2 = 7 \\\alpha = 0.01$

The test hypothesis is

$\\\text{Null Hypothesis }--> H_0: \mu_1 \le \mu_2, or \; H_0: \mu_1 - \mu_2 \le 0 \\\text{Alternate Hypothesis }--> H_1: \mu_1 > \mu_2, or \; H_1: \mu_1 - \mu_2 > 0$

This is a one-sided test because the alternative hypothesis is formulated to detect the difference from the hypothesized mean on the upper side

Now, the value of test static can be found out by following formula:

$\\t_0 = \frac{\bar{X_1} -\bar{X_2} - \Delta_0}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}} \\t_0 = \frac{46.0-35.0 - 0.0}{\sqrt{\frac{16.0}{6}+\frac{25.0}{7}}} \\t_0 = \frac{11.0}{\sqrt{2.6667+4.1667}} \\t_0 = \frac{11.0}{\sqrt{6.2381}} \\t_0 = \frac{11.0}{2.4976} \\t_0 = \textbf{4.40}$

First we calculate degree of freedom
$\\\nu = \frac{\left(\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}\right)^2}{\frac{(S_1^2/n_1)^2}{n_1-1}+\frac{(S_2^2/n_2)^2}{n_2-1}} \\\nu = \frac{\left(\frac{4.0^2}{6} + \frac{5.0^2}{7}\right)^2}{\frac{(4.0^2/6)^2}{6-1}+\frac{(5.0^2/7)^2}{7-1}} \\\nu = \frac{\left(\frac{16.0}{6} + \frac{25.0}{7}\right)^2}{\frac{(16.0/6)^2}{5}+\frac{(25.0/7)^2}{6}} \\\nu = \frac{(2.6667 + 3.5714)^2}{\frac{(2.6667)^2}{5}+\frac{(3.5714)^2}{6}} \\\nu = \frac{(6.2381)^2}{\frac{(7.1111)}{5}+\frac{(12.7551)}{6}} \\\nu = \frac{38.9139}{1.4222+2.1259} \\\nu = \frac{38.9139}{3.5481} \\\nu = 10.9675 \\\nu \approx 11 \\\\\\\text{Using Excel's function }=T.DIST.2T(|t_0|,n-1)\text{, the P-value for }t_0 = 4.4042\text{ in an t-test with 11 degrees of freedom can be computed as P = }P(T_{11}>4.4042)=T.DIST.2T(|4.4042|,11)=\textbf{0.001}.$