The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4 parts per million (ppm). A consumer health group measured the level of the chemical in a random sample of tomatoes obtained from one producer. The levels, in ppm, are shown below.

[tex]\[
\begin{array}{l}
0.310 \quad 0.470 \quad 0.190 \quad 0.720 \quad 0.560 \\
0.910 \quad 0.290 \quad 0.830 \quad 0.490 \quad 0.280 \\
0.310 \quad 0.460 \quad 0.250 \quad 0.340 \quad 0.170 \\
0.580 \quad 0.190 \quad 0.260 \quad 0.470 \quad 0.810 \\
\end{array}
\][/tex]

Does the data provide sufficient evidence to support the claim that the mean level of the chemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm? Use a 0.05 significance level to test the claim that these sample levels come from a population with a mean greater than 0.4 ppm. Use the P-value method of testing hypotheses. Assume that the standard deviation of levels of the chemical in all such tomatoes is 0.21 ppm.

Choose the correct null and alternative hypotheses associated with this experiment.

A. [tex]\(H_0: \mu \neq 0.4\)[/tex] ppm [tex]\(H_1: \mu = 0.4\)[/tex] ppm

B. [tex]\(H_0: \mu = 0.4\)[/tex] ppm [tex]\(H_1: \mu \neq 0.4\)[/tex] ppm

C. [tex]\(H_0: \mu = 0.4\)[/tex] ppm [tex]\(H_1: \mu \ \textgreater \ 0.4\)[/tex] ppm



Answer :

Let's solve this step by step to determine if the data provides sufficient evidence to support the claim that the mean level of the chemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm.

### Step 1: State the Hypotheses

We need to test if the mean level of the chemical in tomatoes is greater than the recommended level of 0.4 ppm. Therefore, the null hypothesis and the alternative hypothesis should be:
- [tex]\( H_0: \mu = 0.4 \)[/tex] ppm (the mean level is equal to 0.4 ppm)
- [tex]\( H_1: \mu > 0.4 \)[/tex] ppm (the mean level is greater than 0.4 ppm)

The correct hypotheses for this experiment are:
[tex]\[ H_0: \mu = 0.4 \text{ ppm} \][/tex]
[tex]\[ H_1: \mu > 0.4 \text{ ppm} \][/tex]

### Step 2: Collect and Summarize the Data

Given sample levels of the chemical in ppm:
[tex]\[ [0.310, 0.470, 0.190, 0.720, 0.56, 0.910, 0.290, 0.830, 0.490, 0.28, 0.310, 0.46, 0.250, 0.340, 0.17, 0.580, 0.190, 0.260, 0.470, 0.81] \][/tex]

The sample mean ([tex]\( \bar{x} \)[/tex]) is 0.4445 ppm.

### Step 3: Determine the Standard Deviation and Sample Size

The given standard deviation of the population is [tex]\( \sigma = 0.21 \)[/tex] ppm.

The sample size ([tex]\( n \)[/tex]) is 20 (as there are 20 data points).

### Step 4: Calculate the Test Statistic

The test statistic for a one-sample Z-test is calculated using the formula:
[tex]\[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \][/tex]

Plugging in the values:
[tex]\[ Z = \frac{0.4445 - 0.4}{0.21 / \sqrt{20}} \approx 0.9477 \][/tex]

### Step 5: Determine the P-Value

Using the Z-score calculated above, we can find the P-value. Since this is a one-tailed test, we are interested in the area to the right of the Z-score.

The P-value associated with a Z-score of 0.9477 is approximately 0.1716.

### Step 6: Make a Decision

Compare the P-value with the significance level [tex]\( \alpha = 0.05 \)[/tex]:

- If [tex]\( P \leq \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( P > \alpha \)[/tex], fail to reject the null hypothesis.

Since [tex]\( 0.1716 > 0.05 \)[/tex], we fail to reject the null hypothesis.

### Conclusion

Based on the sample data and the 0.05 significance level, we do not have sufficient evidence to support the claim that the mean level of the chemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm.