Question

Cadmium, a heavy​ metal, is toxic to animals.​ Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. Some governments have a safety limit for cadmium in dry vegetables at 0.6 parts per million​ (ppm). A hypothesis test is to be performed to decide whether the mean cadmium level in a certain mushroom is less than the​ government's recommended limit. Complete parts​ (a) through​ (c) below.

a) Perform a hypothesis test at the 5% significance level to determine if the mean
cadmium level in the population of Boletus pinicoloa mushrooms is greater than the
government’s recommended limit of 0.5 ppm. Suppose that the standard deviation of
this population’s cadmium levels is o( = 0.37 ppm. Note that the sum of the data is 6.31 ppm. For this problem, be sure to: State your hypotheses, compute your test statistic, give the critical value.

(b) Find the p-value for the test.

Answer 1

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu=0.5\\\\H_a:\mu> 0.5$

The test statistic is z=0.243.

The critical value for a right-tailed test and a significance level of 5% is zc=1.645.

b) P-value = 0.404

Step-by-step explanation:

The question is incomplete:

The sample size is n=12, and the sample data is:

[0.24, 0.92, 0.59, 0.19, 0.62, 0.33, 0.16, 0.25, 0.77, 0.59, 1.33, 0.32 ]

This is a hypothesis test for the population mean.

The claim is that the mean cadmium level in the population of Boletus pinicoloa mushrooms is greater than the government’s recommended limit of 0.5 ppm.

Then, the null and alternative hypothesis are:

$H_0: \mu=0.5\\\\H_a:\mu> 0.5$

The significance level is 0.05.

The sample has a size n=12.

The sample mean is M=0.526.

$M=\dfrac{6.31}{12}=0.526$

The standard deviation of the population is known and has a value of σ=0.37.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.37}{\sqrt{12}}=0.107$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{0.526-0.5}{0.107}=\dfrac{0.026}{0.107}=0.243$

This test is a right-tailed test, so the P-value for this test is calculated as:

$P-value=P(z>0.243)=0.404$

As the P-value (0.404) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean cadmium level in the population of Boletus pinicoloa mushrooms is greater than the government’s recommended limit of 0.5 ppm.

The critical value for a right-tailed test and a significance level of 5% is z=1.645.

As our test statistic z=0.243 is smaller than the critical value, it lies in the acceptance region and the null hypothesis is failed to be rejected.