Answer:
The test statistic for the above hypothesis test of proportion of fish that are infected is 0.472.
Step-by-step explanation:
[tex]\text{consider the provided information}[/tex]
It is given that the total sample space is 50 fish. Out of which we have found that 6 of the fish sampled are infected. Therefore,
n is 50 and x = 6
[tex]\text{The hypotheses is}[/tex]
[tex]H_0: P=0.10, H_a: P<0.10[/tex]
Now, calculate the sample proportion by dividing the infected sample by sample space as shown:
[tex]\hat{p}=\frac{6}{50}=0.12[/tex]
The standard deviation of proportion can be calculated by using the formula:
[tex]\sigma=\sqrt{\frac{p(1-p)}{n}}[/tex]
[tex]\text{Now substitute the respective values in the above formula}[/tex]
[tex]\sigma=\sqrt{\frac{0.10(1-0.10)}{50}}[/tex]
[tex]\sigma=\sqrt{\frac{0.10(0.9)}{50}}[/tex]
[tex]\sigma=\sqrt{0.0018}[/tex]
[tex]\sigma=0.0424[/tex]
[tex]\text{The test statistic is:}[/tex]
[tex]z=\frac{\hat{p}-p}{\sigma}[/tex]
[tex]\text{Now substitute the respective values in the above formula}[/tex]
[tex]z=\frac{0.12-0.10}{0.0424}[/tex]
[tex]z=\frac{0.02}{0.0424}[/tex]
[tex]z=0.472\ approximately [/tex]
Hence, the test statistic for the above hypothesis test of proportion of fish that are infected is 0.472.
