Questions: Some experts believe that 24% of all freshwater fish in a country have such high levels of mercury that they are dangerous to eat. Suppose a fish market has 250 fish tested, and 54 of them have dangerous levels of mercury. Test the hypothesis that this sample is not from a population with 24% dangerous fish, assuming that this is a random sample. Use a significance level of 0.05. Comment on your conclusion. State the null and alternative hypotheses. A. H0: P<0.24 Ha: P>0.24 B. H0: p=0.24 Ha: p>0.24 C. H0: p=0.24 Ha p ≠ 0.24 D. H0: p>0.24 Ha P<0.24 E. H0: p ≠ 0.24 H3: P=0.24

Solution Steps

We are testing the following hypotheses:

• Null Hypothesis: $H_0: p = 0.24$
• Alternative Hypothesis: $H_a: p \neq 0.24$

The sample proportion of fish with dangerous levels of mercury is calculated as: $$\hat{p} = \frac{54}{250} = 0.2160$$

The test statistic $Z$ is calculated using the formula: $$Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$$ Substituting the values: $$Z = \frac{0.2160 - 0.24}{\sqrt{\frac{0.24(1 - 0.24)}{250}}} = -0.8885$$

The P-value associated with the test statistic $Z = -0.8885$ is: $$\text{P-value} = 0.3743$$

For a significance level of $\alpha = 0.05$ in a two-tailed test, the critical region is defined as: $$Z < -1.96 \quad \text{or} \quad Z > 1.96$$

Since the calculated P-value $0.3743$ is greater than $\alpha = 0.05$, we fail to reject the null hypothesis.

Final Answer