a. State the hypotheses and identify the claim.
b. Find the critical value(s).
c. Compute the test value.
d. Make the decision.
e. Summarize the results

1. Warming and Ice Melt The average depth of the
 Hudson Bay is 305 feet. Climatologists were
interested in seeing if warming and ice melt were
affecting the water level. Fifty-five measurements
over a period of randomly selected weeks yielded a
sample mean of 306.2 feet. The population variance
is known to be 3.6. Can it be concluded at the 0.05
level of significance that the average depth has
increased? Is there evidence of what caused this to
happen?

To analyze this problem, let's go through each step systematically:

### a. State the Hypotheses and Identify the Claim
- **Null Hypothesis (\(H_0\))**: The average depth of Hudson Bay has not increased, i.e., \(\mu = 305\) feet.
- **Alternative Hypothesis (\(H_a\))**: The average depth of Hudson Bay has increased, i.e., \(\mu > 305\) feet.

**Claim**: The claim is that the average depth of Hudson Bay has increased due to warming and ice melt.

### b. Find the Critical Value(s)
Since we are conducting a one-tailed test at the 0.05 significance level, we will find the critical z-value for a right-tailed test. Using the standard normal distribution:

- **Critical Value (z)**: For \(\alpha = 0.05\), the critical z-value is approximately **1.645**.

### c. Compute the Test Value
To compute the test statistic (z), we use the formula:

\[
z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}
\]

Where:
- \(\bar{x} = 306.2\) (sample mean)
- \(\mu = 305\) (population mean)
- \(\sigma^2 = 3.6\) (population variance) → \(\sigma = \sqrt{3.6} \approx 1.897\)
- \(n = 55\) (sample size)

Now substituting these values into the formula:

\[
z = \frac{306.2 - 305}{1.897/\sqrt{55}}
\]

Calculating the standard error:

\[
SE = \frac{1.897}{\sqrt{55}} \approx 0.255
\]

Now compute the z-value:

\[
z = \frac{306.2 - 305}{0.255} \approx \frac{1.2}{0.255} \approx 4.71
\]

### d. Make the Decision
- Compare the computed z-value (4.71) to the critical value (1.645).
- Since \(4.71 > 1.645\), we reject the null hypothesis.

### e. Summarize the Results
At the 0.05 level of significance, there is sufficient evidence to conclude that the average depth of Hudson Bay has increased, suggesting that warming and ice melt may be affecting the water level.

---

### Follow-Up Questions
1. What are the implications of an increase in the average depth of Hudson Bay?
2. How would the results change if the sample size were smaller?
3. What other factors could contribute to changes in water levels besides warming and ice melt?
4. How can we further investigate the cause of the increase in depth?
5. What would be the steps for a two-tailed test in this scenario?

### Tip
When conducting hypothesis tests, always ensure to clearly define your hypotheses before calculating test statistics to avoid any confusion during the analysis.

Can it be concluded at the 0.05 level of significance that the average depth of Hudson Bay has increased? Is there evidence of what caused this to happen?

Explore the hypothesis testing process for assessing if the average depth of Hudson Bay has increased due to warming and ice melt. This detailed analysis includes key statistical concepts, critical values, and test statistics based on a sample of 55 measurements. Ideal for high school students in Grades 11-12.

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