The following output from MINITAB presents the results of a hypothesis test.
Test of $m u = 36$ vs. not $= 36$
The assumed standard deviation $12.1$

$\begin{array}{lccccl}
\text { N } & \text { Mean } & \text { SE Mean } & 95 \% \text { CI } & Z & \text { P } \
52 & 39.08 & 1.970028 & (1.55816684,9.280675) & 1.835553 & 0.066424
\end{array}$
What are the null and alternate hypotheses?
A) $H _ { 0 } : \mu \neq 36 , H _ { 1 } : \mu = 36$
B) $H _ { 0 } : \mu = 36 , H _ { 1 } : \mu = 39.08$
C) $H _ { 0 } : \mu = 39.08 , H _ { 1 } : \mu \neq 39.08$
D) $H _ { 0 } : \mu = 36 , H _ { 1 } : \mu \neq 36$

Question

The following output from MINITAB presents the results of a hypothesis test.
Test of $m u = 36$ vs. not $= 36$
The assumed standard deviation $12.1$

$\begin{array}{lccccl}
\text { N } & \text { Mean } & \text { SE Mean } & 95 \% \text { CI } & Z & \text { P } \
52 & 39.08 & 1.970028 & (1.55816684,9.280675) & 1.835553 & 0.066424
\end{array}$
What are the null and alternate hypotheses?
A) $H _ { 0 } : \mu \neq 36 , H _ { 1 } : \mu = 36$
B) $H _ { 0 } : \mu = 36 , H _ { 1 } : \mu = 39.08$
C) $H _ { 0 } : \mu = 39.08 , H _ { 1 } : \mu \neq 39.08$
D) $H _ { 0 } : \mu = 36 , H _ { 1 } : \mu \neq 36$

Quizplus · Accepted Answer

The null hypothesis always specifies a value for the parameter that is tested without an inequality or not equal sign, and the alternative hypothesis is what the test is trying to provide evidence for. In this case, the test is for $\mu = 36$ versus $\mu \neq 36$, making D the correct choice.

The Following Output from MINITAB Presents the Results of a Hypothesis