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) .2302
B) .1151
C) .3849
D) .6151

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) .2302
B) .1151
C) .3849
D) .6151

Quizplus · Accepted Answer

The two-tailed p-value for a z-score of -1.20 is found by looking up the z-score in a standard normal distribution table, which gives approximately 0.1151 for one tail. Doubling this value for a two-tailed test gives 0.2302.

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