Consider the following set of salary data: $\begin{array}{lrr}
\hline & \text { Men (1) } & \text { Women (2) } \
\hline \text { Sample Size } & 100 & 80 \
\text { Mean } & \$ 12,850 & \$ 13,000 \
\text { Standard Deviation } & \$ 345 & \$ 500 \
\hline
\end{array}$

Suppose the test statistic turned out to be $z = - 1.20$ (not the correct value). Find a two-tailed $p$-value for this test statistic.
A) .6151
B). 3849
C). 1151
D) .2302

Question

Consider the following set of salary data: $\begin{array}{lrr}
\hline & \text { Men (1) } & \text { Women (2) } \
\hline \text { Sample Size } & 100 & 80 \
\text { Mean } & \$ 12,850 & \$ 13,000 \
\text { Standard Deviation } & \$ 345 & \$ 500 \
\hline
\end{array}$

Suppose the test statistic turned out to be $z = - 1.20$ (not the correct value). Find a two-tailed $p$-value for this test statistic.
A) .6151
B). 3849
C). 1151
D) .2302

Quizplus · Accepted Answer

The two-tailed $p$-value for a $z$-score of $-1.20$ is found by looking up the absolute value of the $z$-score in a standard normal distribution table or using a calculator. The area to the right of $z = 1.20$ is approximately 0.1151, so for a two-tailed test, you double this value to get $2 \times 0.1151 = 0.2302$.

Consider the Following Set of Salary Data Suppose the Test Statistic Turned Out to Be