Find the standardized test statistic, t, to test the hypothesis that μ1 = μ2. Two samples are randomly selectedand $$\begin{array} { l l }
\mathrm { n } _ { 1 } = 14 & \mathrm { n } _ { 2 } = 12 \
\overline { \mathrm { x } } 1 = 5 & \overline { \mathrm { x } } 2 = 6 \
\mathrm {~s} _ { 1 } = 2.5 & \mathrm {~s} _ { 2 } = 2.8
\end{array}$$ come from populations that are normal. The sample statistics are given below.
A) -0.954
B) -0.915
C) -1.558
D) -0.909

Question

Find the standardized test statistic, t, to test the hypothesis that μ1 = μ2. Two samples are randomly selectedand $$\begin{array} { l l } 
\mathrm { n } _ { 1 } = 14 & \mathrm { n } _ { 2 } = 12 \
\overline { \mathrm { x } } 1 = 5 & \overline { \mathrm { x } } 2 = 6 \
\mathrm {~s} _ { 1 } = 2.5 & \mathrm {~s} _ { 2 } = 2.8
\end{array}$$ come from populations that are normal. The sample statistics are given below.
A) -0.954
B) -0.915
C) -1.558
D) -0.909

Quizplus · Accepted Answer

The standardized test statistic, $t$, can be calculated using the formula for two independent samples:$$ t = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$Plugging in the given values:$$ t = \frac{5 - 6}{\sqrt{\frac{2.5^2}{14} + \frac{2.8^2}{12}}} = \frac{-1}{\sqrt{\frac{6.25}{14} + \frac{7.84}{12}}} = \frac{-1}{\sqrt{0.446 + 0.653}} = \frac{-1}{\sqrt{1.099}} = \frac{-1}{1.048} = -0.954$$

Find the Standardized Test Statistic, T, to Test the Hypothesis