Hypothesis Test for Difference in Population Means ( $σ$ Unknown)
You wish to test the following claim $(H_{a})$ at a significance level of $α = 0.10$
$\begin{array}{l} H_{o} : μ_{1} = μ_{2} \\ H_{a} : μ_{1} \neq μ_{2} \end{array}$
You believe both populations are normally distributed, but you do not know either. We will assume that the population variances are not equal.

You obtain a sample of size $n_{1} = 14$ with a mean of $M_{1} = 68.6$ and a stand from the first population. You obtain a sample of size $n_{2} = 19$ with a mean c deviation of $S D_{2} = 16.5$ from the second population.

What is the test statistic for this sample? (Report answer accurate to three di test statistic $=$

What is the $p$ -value for this sample? For this calculation, use the conservative degrees of freedom. The degrees of freedom is the minimum of $n_{1} - 1$ and $n_{2}$ to four decimal places.)
$p -value =$
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4. Calculate degrees of freedom: $d f = min (n_{1} - 1, n_{2} - 1) = min (13, 18) = 13$

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Step 1 ：1. Compute the sample mean difference: $\bar{d} = M_{1} - M_{2} = 68.6 - 81 = - 12.4$

Step 2 ：2. Calculate standard error: $S E = \sqrt{\frac{S_{1}^{2}}{n_{1}} + \frac{S_{2}^{2}}{n_{2}}} = \sqrt{\frac{{23.1}^{2}}{14} + \frac{{16.5}^{2}}{19}} = 7.7467$

Step 3 ：3. Compute the test statistic: $t = \frac{\bar{d} - 0}{S E} = \frac{- 12.4}{7.7467} = - 1.6019$

Step 4 ：4. Calculate degrees of freedom: $d f = min (n_{1} - 1, n_{2} - 1) = min (13, 18) = 13$

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