The highest acceptable level of pesticide found in quail has been limited to 0.35 parts per million (ppm). A hunting organization measured the level of the pesticide found in a random sample of 20 quail harvested from field trials. The sample produced a sample mean of .44. Assume that the population standard deviation is 0.21 ppm. Using alpha [] = 0.05, does the data provide sufficient evidence to conclude that the mean level of pesticide is greater that the limit of 0.35 ppm?
2a. Use the critical value z0 method to test the hypothesis.
(References: example 7 through 10 pages 385 - 388, end of section exercises 39 – 44 pages 392 - 393) (6 points)
H0: u≤0.35
Ha: u>0.35
Critical z0:
Decision:
Interpretation:
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion
To test the hypothesis using the critical value z0 method, follow these steps:
Step 1: State the null and alternative hypotheses.
H0: The mean level of pesticide (μ) is less than or equal to 0.35 ppm.
Ha: The mean level of pesticide (μ) is greater than 0.35 ppm.
Step 2: Determine the significance level (α).
Given that α = 0.05, or a 5% level of significance.
Step 3: Determine the test statistic.
The test statistic for testing the population mean is the z-score. The formula for the z-score is:
z = (x̄ - μ) / (σ / √n)
Where:
x̄ is the sample mean (0.44 ppm),
μ is the hypothesized population mean (0.35 ppm),
σ is the population standard deviation (0.21 ppm), and
n is the sample size (20).
Plugging in the values, we get:
z = (0.44 - 0.35) / (0.21 / √20)
z = 2.121
Step 4: Determine the critical value.
Since the alternative hypothesis is one-tailed and we want to know if the mean is greater than 0.35 ppm, we need to find the critical value z0 for a right-tailed test with a significance level of 0.05. This can be done using a z-table or calculator.
The critical value z0 for a right-tailed test with α = 0.05 is approximately 1.645.
Step 5: Make a decision.
If the test statistic (z) is greater than the critical value (z0), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, z = 2.121 is greater than z0 = 1.645, so we reject the null hypothesis.
Step 6: Interpret the result.
Based on the data, there is sufficient evidence to conclude that the mean level of pesticide is greater than the limit of 0.35 ppm.