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 350 fish tested, and 80 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. State the null and alternative hypothesis, determine the z-test statistic, find the p-value, does the conclusion support or reject the null.

Step-by-step explanation:

Hello!

The study variable is X: the number of freshwater fish that contain dangerous levels of mercury.

It is known that the population proportion is believed to be 0,24

The investigator hypothesizes that the population proportion is not from aa population with 24% dangerous fish. Symbolically: ρ ≠ 0.24 this is going to represent the alternative hypothesis, and in the null hypothesis represents the already known information about the freshwater fish.

H₀: ρ = 0.24

H₁: ρ ≠ 0.24

α = 0.05

The statistic you have to use is an approximate Z (since the sample is large enough, n350, you can apply the central limit theorem to make the approximation.)

Z = (ρ₀ - ρ) / √ ((p (1-p)) / n) ≈ N (0; 1)

Z = (0.23-0.24) / √ ((0.24 (1-0.24)) / 350) = - 0.43

ρ₀ is the sample proportion = x/n = 80/350 = 0.23

P-value is two-tailed = 0.6671

To decide using the p-value you have to compare it with the signification level. If its bigger than alpha then you do not reject the null hypothesis, but if it is less or equal to the significance level, you reject the null hypothesis.

In this case, the decision is to not reject the null hypothesis. So you can say that there is enough evidence to say that the sampled fishes are from the population with 24% of mercury.

I hope you have a SUPER day!