Consider the following set of salary data: $$\begin{array} { l r r }
\hline & \text { Men (1) } & \text { Women (2) } \
\hline \text { Sample Size } & 100 & 80 \
\text { Mean } & \$ 12,850 & \$ 13,000 \
\text { Standard Deviation } & \$ 345 & \$ 500 \
\hline
\end{array}$$ 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) .3849
C) .1151
D) .6151

Question

Consider the following set of salary data: $$\begin{array} { l r r } 
\hline & \text { Men (1) } & \text { Women (2) } \
\hline \text { Sample Size } & 100 & 80 \
\text { Mean } & \$ 12,850 & \$ 13,000 \
\text { Standard Deviation } & \$ 345 & \$ 500 \
\hline
\end{array}$$ 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) .3849
C) .1151
D) .6151

Quizplus · Accepted Answer

The correct answer is A. The p-value for a two-tailed test with z = -1.20 is 0.2302. This can be found using a standard normal distribution table or a calculator.

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