Question

The FDA regulates that a fish that is consumed is allowed to contain at most 1 mg/kg of mercury. In Florida, bass fish were collected in 53 different lakes to measure the amount of mercury in the fish from each of the 53 lakes. Do the data provide enough evidence to show that the fish in all Florida lakes have different mercury than the allowable amount?
Required:
State the random variable, population parameter, and hypotheses.

Answer 1

Answer:

Yes. At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.

The random variable is the sample mean amount of mercury in the bass fish from the lakes of Florida.

The population parameter is the mean amount of mercury in the bass fish of Florida lakes.

The alternative hypothesis (Ha) states that the amount of mercury significantly differs from 1 mg/kg.

The null hypothesis (H0) states that the amount of mercury is not significantly different from 1 mg/kg.

$H_0: \mu=1\\\\H_a:\mu\neq 1$

Step-by-step explanation:

The question is incomplete.

There is no data provided.

We will work with a sample mean of 0.95 mg/kg and sample standard deviation of 0.15 mg/kg to show the procedure.

This is a hypothesis test for the population mean.

The claim is that the fish in all Florida lakes have different mercury than the allowable amount (1 mg of mercury per kg of fish).

Then, the null and alternative hypothesis are:

$H_0: \mu=1\\\\H_a:\mu\neq 1$

The significance level is assumed to be 0.05.

The sample has a size n=53.

The sample mean is M=0.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.15.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.15}{\sqrt{53}}=0.0206$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.95-1}{0.0206}=\dfrac{-0.05}{0.0206}=-2.427$

The degrees of freedom for this sample size are:

$df=n-1=53-1=52$

This test is a two-tailed test, with 52 degrees of freedom and t=-2.427, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t$

As the P-value (0.019) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the fish in all Florida lakes have different mercury than the allowable amount.