Question

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

Answer 1

Given

`\begin{gathered} H_0\colon p=0.61 \\ H_1\colon p<0.61 \end{gathered}`

The test statistics formula for proportion is given by

`z=\frac{p-p_0}{\sqrt[]{(p_0(1-p_0))/(n)}}`

The parameters are

`\begin{gathered} p=(61)/(97)=0.6289 \\ p_0=0.61 \\ n=97 \end{gathered}`

using the formula

`\begin{gathered} z=\frac{0.6289-0.61}{\sqrt[]{(0.61(1-0.61))/(97)}} \\ z=\frac{0.0189}{\sqrt[]{0.002453}} \\ z=0.3816 \end{gathered}`

Thus, the test statistics to 3 decimal place is 0.382