A random sample of 15 chemists from Washington state shows an average salary of $45720 with a standard deviation of $654. A random sample of 24 chemists from Florida state shows an average salary of $42403 with a standard deviation of $665. A chemist that has worked in both states believes that chemists in Washington make a different amount than chemists in Florida. At a=0.01 is this chemist correct? Let Washington be sample 1 and Florida be sample 2. The correct hypotheses are: O Ho: H1 < H2 HA: 41 > µ2(claim) O Ho: µ1 2 H2 HA: H1 < µ2(claim) O Ho: µ1 = µ2 HA: H1 + µ2(claim) Since the level of significance is 0.01 the critical value is 2.749 and -2.749 The test statistic is: |(round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho

To test
$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_1:{\mu }_1\ne {\mu }_2 \end{gathered}$$

The pooled standard deviation is
$\begin{array}{rcl}{s}_p& =& \sqrt{\frac{\left(n_1-1\right){s_1}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}}\\ & =& \sqrt{\frac{\left(15-1\right){\left(654\right)}^2+\left(24-1\right){\left(665\right)}^2}{15+24-2}}\\ & =& \sqrt{\frac{5988024+10171175}{37}}\\ & =& \sqrt{436735.108}\\ & =& 660.8594\end{array}$

Then, the test statistic is
$\begin{array}{rcl}t& =& \frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{{\displaystyle \frac{1}{n_1}}+{\displaystyle \frac{1}{n_2}}}}\\ & =& \frac{45720-42403}{660.8594\sqrt{{\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{24}}}}\\ & =& \frac{3317}{660.8594\left(0.3291\right)}\\ & =& \frac{3317}{217.5154}\\ & =& 15.249\end{array}$

The p-value is 0.