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 $\sigma _ { 1 } ^ { 2 } \neq \sigma _ { 2 }^ { 2 } \text {. }$ $$\begin{array} { l l }
\mathrm { n } _ { 1 } = 25 & \mathrm { n } _ { 2 } = 30 \
\overline { \mathrm { x } } 1 = 33 & \overline { \mathrm { x } } 2 = 31 \
\mathrm {~s} 1 = 1.5 & \mathrm {~s} _ { 2 } = 1.9
\end{array}$$
A)4.361
B)3.287
C)2.892
D)1.986

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 $\sigma _ { 1 } ^ { 2 } \neq \sigma _ { 2 }^ { 2 } \text {. }$ $$\begin{array} { l l } 
\mathrm { n } _ { 1 } = 25 & \mathrm { n } _ { 2 } = 30 \
\overline { \mathrm { x } } 1 = 33 & \overline { \mathrm { x } } 2 = 31 \
\mathrm {~s} 1 = 1.5 & \mathrm {~s} _ { 2 } = 1.9
\end{array}$$
A)4.361
B)3.287
C)2.892
D)1.986

Quizplus · Accepted Answer

The correct choice is A because the standardized test statistic, t, is calculated as:$$t = \frac{(\overline{x}_1 - \overline{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}$$Since we are assuming that $\sigma _ { 1 } ^ { 2 } \neq \sigma _ { 2 }^ { 2 } \text {,}$ we use the following formula:$$t = \frac{(\overline{x}_1 - \overline{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$Plugging in the given values, we get:$$t = \frac{(33 - 31)}{\sqrt{\frac{1.5^2}{25} + \frac{1.9^2}{30}}} = 4.361$$Therefore, the standardized test statistic, t, is 4.361.

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 $\mu _ { 1 } = \mu _ { 2 } .$