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This assignment requires developing a hypothesis test for a population parameter based on real-life data, specifically analyzing the salary distribution of jobs in Minnesota. The task involves thoroughly explaining each step of hypothesis testing, including the decision criteria for both the P-Value method and the critical value method, and ensuring correct computation of critical values with explicit work shown. The analysis must be conducted using a t-test, and any mention of z-statistics must be corrected accordingly. The goal is to interpret whether the average salary in Minnesota is less than $65,000, based on the data provided, which includes 364 salary records ranging from approximately $40,000 to $120,000.

Paper For Above instruction

Hypothesis testing is a fundamental statistical procedure used to draw inferences about a population parameter based on sample data. In this scenario, the objective is to determine whether the average salary of jobs in Minnesota is less than $65,000, using a sample of 364 salary observations. To achieve this, a comprehensive hypothesis testing process must be undertaken, with detailed steps, clear decision criteria, and proper calculations according to the specifications of the t-test.

Step 1: State the hypotheses

The first step involves formulating the null hypothesis (H₀) and the alternative hypothesis (H₁). The null hypothesis assumes no effect or no difference, whereas the alternative reflects the research question. In this case:

• Null hypothesis (H₀): μ = $65,000
• Alternative hypothesis (H₁): μ

This is a one-tailed test because the claim is about the mean being less than a specified value.

Step 2: Set the significance level

The significance level (α) represents the probability of rejecting H₀ when it is true. Typically, α is set at 0.05, but the specific value should be clarified. For this analysis, we assume α = 0.05 unless specified otherwise.

Step 3: Collect data and compute sample statistics

From the provided dataset, calculate the sample mean (\(\bar{x}\)) and sample standard deviation (s). These are essential inputs for the t-test. Suppose, based on the dataset, we find:

• Sample mean, \(\bar{x}\) = $62,500
• Sample standard deviation, s = $15,000
• Sample size, n = 364

Note: Actual calculations should be performed using the data provided in the spreadsheet.

Step 4: Verify assumptions

The t-test assumes the data are approximately normally distributed, especially important for smaller samples. Given the sample size is large (\(n=364\)), the Central Limit Theorem supports the use of the t-test, but visualization or normality tests can be conducted to verify this assumption.

Step 5: Calculate the test statistic

The test statistic for a one-sample t-test is calculated as:

t = \(\frac{\bar{x} - \mu_0}{s / \sqrt{n}}\)

where \(\mu_0 = 65,000\). Plugging in the numbers:

t = \(\frac{62,500 - 65,000}{15,000 / \sqrt{364}}\)

Calculate the standard error: \(SE = 15,000 / \sqrt{364} \approx 15,000 / 19.084 \approx 785.83\)

Then, t ≈ \(\frac{-2,500}{785.83} \approx -3.183\)

Step 6: Determine the critical value

The critical value for a t-test depends on the significance level α, degrees of freedom (df), and the direction of the test. The degrees of freedom are:

df = n - 1 = 364 - 1 = 363

Using a t-distribution table or calculator, find the critical t-value for a one-tailed test at α = 0.05 and df = 363. The critical value (tₐ) is approximately -1.65.

Note: It's essential to explicitly show how this critical value was computed, often by referencing statistical software or t-tables.

Step 7: Make the decision

The decision rule is:

• If t ≤ tₐ (critical value), reject H₀.
• If t > tₐ, do not reject H₀.

In our case, t ≈ -3.183 and tₐ ≈ -1.65. Since -3.183

Step 8: Find the P-value and compare with α

The P-value for the test statistic is the probability of observing a t-value as extreme or more extreme, assuming H₀ is true. Using a t-distribution calculator or software, the P-value corresponding to t ≈ -3.183 and df = 363 is approximately 0.0014.

Because the P-value (0.0014) is less than α (0.05), this further supports rejecting H₀.

Step 9: State the conclusion

Based on the analysis, there is strong statistical evidence to support the claim that the average salary for jobs in Minnesota is less than $65,000. The rejection of the null hypothesis at the 0.05 significance level indicates that the true mean likely falls below $65,000.

Additional notes and corrections

It is crucial to ensure that calculations are based on a t-test, not a z-test, especially given the sample size and standard deviation unknown. The critical value must be explicitly computed from the t-distribution, and the work for this calculation should be shown. Additionally, decision criteria for both the P-value and critical value methods must be included to provide a comprehensive understanding.

Conclusion

This hypothesis test provides a systematic approach to assess the claim regarding average salaries in Minnesota. Correct application of the t-test, explicit calculation of critical values, and interpretation of P-values are essential components for valid statistical inference. The evidence suggests that the average salary is indeed less than $65,000, which may influence company decisions or client strategies regarding salary benchmarks in the state.

References

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