Answer :
Solution
We have the proportion of the freshwater fish is
[tex]\begin{gathered} p_0=23\% \\ p_0=0.23 \end{gathered}[/tex]Now, from the sample we have the proportion
[tex]\begin{gathered} p=\frac{48}{200} \\ p=0.24 \end{gathered}[/tex]To test the hypothesis, we have
[tex]\begin{gathered} H_0\colon p_0=0.23 \\ H_1\colon p_0\ne0.23 \end{gathered}[/tex]We have
[tex]\alpha=0.05,n=200[/tex]The test statistics
[tex]z=\frac{p-p_0}{\sqrt[]{\frac{p_0(1-p_0)}{n}}}[/tex]Substituting the parameters, we obtain
[tex]\begin{gathered} z=\frac{p-p_0}{\sqrt[]{\frac{p_0(1-p_0)}{n}}} \\ z=\frac{0.24-0.23}{\sqrt[]{\frac{0.23(1-0.23)}{200}}} \\ z=\frac{0.01}{\sqrt[]{\frac{0.23\times0.77}{200}}} \\ z=0.3360514064 \end{gathered}[/tex]Therefore, the test statistics is
[tex]z=0.336\text{ (to thr}ee\text{ decimal places)}[/tex]To find the P - value
Using the table, the p - value for the z score of 0.336 is
[tex]\begin{gathered} p-value=0.736871 \\ p-value=0.7369\text{ (to four decimal places)} \end{gathered}[/tex]Notice that
[tex]p-value=0.7369>0.05=\alpha[/tex]Therefore, since the p -value is greater than the level of significance, we fail to reject the null hypothesis, That is we support the null hypothesis