Let's go through the hypothesis test step by step.

Step 1: Write Down the Null and Alternative Hypothesis Using Words

• Null Hypothesis (H₀): The mean phosphorus amount in the stream is equal to or less than the upper limit of 0.05 mg/L. This suggests that the stream does not have a phosphorus level exceeding the natural limit.

• Alternative Hypothesis (H₁): The mean phosphorus amount in the stream is greater than the upper limit of 0.05 mg/L. This suggests that the stream has a phosphorus level exceeding the natural limit.

Step 2: Write Down the Null and Alternative Hypothesis Using Symbols

• Null Hypothesis (H₀): $\mu \leq 0.05$ mg/L
• Alternative Hypothesis (H₁): $\mu > 0.05$ mg/L

Step 3: What is the Value of $\alpha$ in This Hypothesis Test?

• The significance level $\alpha$ is given as 0.05.

Step 4: Determine an Appropriate Distribution (Standard Normal Distribution or t-Distribution) to Use

• t-Distribution: Since the sample size is small ($n = 16$), and we do not know the population standard deviation, the t-distribution is more appropriate.

Step 5: Sketch the Distribution and Shade the Rejection Region

The t-distribution is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. You would shade the right tail of the distribution corresponding to the rejection region, as this is a one-tailed test.

Step 6: Calculate the Value of the Test Statistic

The test statistic for a one-sample t-test is calculated using the formula:

$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{x} = 0.053$ mg/L (sample mean)
• $\mu_0 = 0.05$ mg/L (hypothesized population mean)
• $s = 0.008$ mg/L (sample standard deviation)
• $n = 16$ (sample size)

Now let's calculate it:

$t = \frac{0.053 - 0.05}{\frac{0.008}{\sqrt{16}}} = \frac{0.003}{0.002} = 1.5$

So, the test statistic $t$ is 1.5.

Step 7: Determine the p-value Using an Appropriate Table or Technology

To find the p-value, we refer to the t-distribution table or use a calculator. The degrees of freedom (df) is $n - 1 = 16 - 1 = 15$.

Looking up the t-value of 1.5 with 15 degrees of freedom, the p-value is around 0.08.

Step 8: Compare the p-value and the Value of $\alpha$, and Determine Whether You Should Reject or Not Reject the Null Hypothesis

• The p-value $\approx 0.08$ is greater than the significance level $\alpha = 0.05$.
• Since the p-value is not less than $\alpha$, we do not reject the null hypothesis.

Step 9: State the Conclusion of the Hypothesis Testing

• Conclusion: Based on the sample data, there is insufficient evidence at the 0.05 significance level to conclude that the mean phosphorus level in the stream is greater than the upper limit of 0.05 mg/L.

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Would you like further details or have any questions?

Related Questions:

1. How does sample size affect the choice between using a t-distribution and a normal distribution?
2. What assumptions must be met for the t-test to be valid in this context?
3. How would the conclusion change if the sample size were larger, say 50?
4. What role does the significance level $\alpha$ play in hypothesis testing?
5. Can we use this test for any other water quality parameters besides phosphorus?
6. What are some potential consequences of a Type I error in this context?
7. How would you interpret the results if the p-value was exactly 0.05?
8. What would happen if the sample mean was exactly 0.05 mg/L?

Tip:

When performing hypothesis tests, always ensure that the assumptions underlying the test are satisfied before drawing conclusions. This is particularly important when working with small sample sizes.