Question

Cadmium, a heavy​ metal, is toxic to animals.​ Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. A government set a safety limit for cadmium in dry vegetables at 0.5 parts per million​ (ppm). The cadmium levels in a random sample of one species of edible mushroom are in the accompanying data set. At the ​5% significance​ level, do the data provide sufficient evidence to conclude that the mean cadmium level in this species of mushroom is greater than the​ government's recommended limit of 0.5 ​ppm? Assume that the population standard deviation of cadmium levels in this species of mushroom is 0.42 ppm. Preliminary data analyses indicate that applying the​ z-test is reasonable.​ (Note: The sum of the data is 6.21 ​ppm.)

Compute the value of the test statistic.

Answer 1

The value of the test statistic is 1.25.

The step-by-step solution to determine the value of the test statistic:

Step 1: Formulate the null and alternative hypotheses

The null hypothesis (H₀) is that the mean cadmium level in this species of mushroom is equal to the government's recommended limit of 0.5 ppm.

H₀: μ = 0.5 ppm

The alternative hypothesis (H₁₂) is that the mean cadmium level in this species of mushroom is greater than 0.5 ppm.

H₁: μ > 0.5 ppm

Step 2: Calculate the sample mean (X) and sample standard deviation (s)

Given the data set, we can calculate the sample mean (X) as follows:

X = ΣXi/n = 6.21 ppm / 6 = 1.035 ppm

We are also given that the population standard deviation (σ) is 0.42 ppm. Since we are assuming that the population standard deviation is known, we can use the formula for the z-test statistic.

Step 3: Calculate the z-test statistic

The z-test statistic is calculated as follows:

z = (X - μ₀) / (σ/√n)

where:

X is the sample mean (1.035 ppm)

μ₀ is the hypothesized mean (0.5 ppm)

σ is the population standard deviation (0.42 ppm)

n is the sample size (6)

Plugging in the values, we get:

z = (1.035 ppm - 0.5 ppm) / (0.42 ppm/√6)

z = 1.25

Therefore, the value of the test statistic is 1.25.