Find the standardized test statistic, t, to test the claim that $\mu _ { 1 } = \mu _ { 2 }$ Two samples are random, independent, and come from populations that are normally distributed. The sample statistics are given below. Assume that $\sigma \sigma _ { 1 } ^ { 2 } = \sigma _ { 2 } ^ { 2 }$ $$\begin{array} { l l }
\mathrm { n } _ { 1 } = 14 & \mathrm { n } _ { 2 } = 12 \
\overline { \mathrm { x } _ { 1 } } = 14 & \overline { \mathrm { x } 2 } = 15 \
\mathrm {~s} _ { 1 } = 2.5 & \mathrm {~s} _ { 2 } = 2.8
\end{array}$$
A)-0.962
B)-0.813
C)-1.326
D)-1.101

Question

Find the standardized test statistic, t, to test the claim that $\mu _ { 1 } = \mu _ { 2 }$ Two samples are random, independent, and come from populations that are normally distributed. The sample statistics are given below. Assume that $\sigma \sigma _ { 1 } ^ { 2 } = \sigma _ { 2 } ^ { 2 }$ $$\begin{array} { l l } 
\mathrm { n } _ { 1 } = 14 & \mathrm { n } _ { 2 } = 12 \
\overline { \mathrm { x } _ { 1 } } = 14 & \overline { \mathrm { x } 2 } = 15 \
\mathrm {~s} _ { 1 } = 2.5 & \mathrm {~s} _ { 2 } = 2.8
\end{array}$$
A)-0.962
B)-0.813
C)-1.326
D)-1.101

Quizplus · Accepted Answer

The standardized test statistic, $t$, can be calculated using the formula for two independent samples assuming equal variances: $$t = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{s_p^2(\frac{1}{n_1} + \frac{1}{n_2})}}$$ where $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$ is the pooled variance. Plugging in the given values: $n_1 = 14, n_2 = 12, \overline{x}_1 = 14, \overline{x}_2 = 15, s_1 = 2.5, s_2 = 2.8$, we find $s_p^2 = \frac{(14-1)(2.5)^2 + (12-1)(2.8)^2}{14 + 12 - 2}$ and then calculate $t$. The calculation yields a $t$ value of approximately -0.962.

Find the Standardized Test Statistic, T, to Test the Claim μ1=μ2\mu _ { 1 } = \mu _ { 2 }μ1​=μ2​

Find the Standardized Test Statistic, T, to Test the Claim $\mu _ { 1 } = \mu _ { 2 }$