A large fishing farm with thousands of fish has been treating its fish to stop a spreading fungal infection. The owner of the fishing farm claims that fewer than 10% of the fish are infected. A random sample of 50 fish is taken to determine the proportion p that is infected in this population. A careful examination determines that 6 of the fish sampled are infected. The test statistic for the above hypothesis test of proportion of fish that are infected is... (Round your answer to two decimal places.)

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.