Question

Hi, can you help me answer this question please, thank you

Answer 1

The test you are running is:

`\begin{gathered} H_0\colon p=0.55 \\ H_a\colon p>0.55 \end{gathered}`

Then, you have a right-tailed p-value, then:

`\begin{gathered} p-value=P(Z>z_0) \\ p-value=1-P(Z\le z_0) \end{gathered}`

By replacing p-value =0.55:

`\begin{gathered} 0.55=1-P(Z\le z_0) \\ P(Z\le z_0)=1-0.55 \\ P(Z\le z_0)=0.45 \end{gathered}`

Then, in a standard normal table look for the corresponding z:

As you can see, z=-0.12 has a p-value=0.4522 and z=-0.13 has a p-value=0.4483.

Then, the average will be:

`\begin{gathered} (0.4522+0.4483)/(2)=0.4502 \\ \text{Then the z-score is:} \\ (-0.12+(-0.13))/(2)=-0.125 \end{gathered}`

Answer: the test statistic z=-0.125