Questions: A recent survey identified the top accounting firms within 10 geographical regions across country X. The top 2 regions reported a combined growth of 19% and 18%. A characteristic description of the accounting firms in these two regions included the number of partners in the firms. Attached below is a sample of the number of partners for 20 firms for each region Complete (a) through (c) below a. At the 0.05 level of significance when pooling the variances, is there evidence of a difference between the two regions' accounting firms with respect to the mean number of partners? Let μ1 be the mean number of partners for the highest growth region and μ2 be the mean number of partners for the second highest growth region Determine the hypotheses Choose the correct answer below. A. H0 μ1 ≤ μ2 B. H0 μ1 ≥ μ2 H1 · μ1>μ2 H1 · μ1<μ2 C. H0 μ1=μ2 D. H0 μ1 ≠ μ2 H1 μ1 ≠ μ2

A recent survey identified the top accounting firms within 10 geographical regions across country X. The top 2 regions reported a combined growth of 19% and 18%. A characteristic description of the accounting firms in these two regions included the number of partners in the firms. Attached below is a sample of the number of partners for 20 firms for each region Complete (a) through (c) below a. At the 0.05 level of significance when pooling the variances, is there evidence of a difference between the two regions' accounting firms with respect to the mean number of partners? Let μ1 be the mean number of partners for the highest growth region and μ2 be the mean number of partners for the second highest growth region Determine the hypotheses Choose the correct answer below. A. H0 μ1 ≤ μ2 B. H0 μ1 ≥ μ2 H1 · μ1>μ2 H1 · μ1<μ2 C. H0 μ1=μ2 D. H0 μ1 ≠ μ2 H1 μ1 ≠ μ2
Transcript text: A recent survey identified the top accounting firms within 10 geographical regions across country X. The top 2 regions reported a combined growth of $19\%$ and $18\%$. A characteristic description of the accounting firms in these two regions included the number of partners in the firms. Attached below is a sample of the number of partners for 20 firms for each region Complete (a) through (c) below a. At the 0.05 level of significance when pooling the variances, is there evidence of a difference between the two regions' accounting firms with respect to the mean number of partners? Let $\mu_{1}$ be the mean number of partners for the highest growth region and $\mu_{2}$ be the mean number of partners for the second highest growth region Determine the hypotheses Choose the correct answer below. A. $\mathrm{H}_{0} \mu_{1} \leq \mu_{2}$ B. $H_{0} \mu_{1} \geq \mu_{2}$ $H_{1} \cdot \mu_{1}>\mu_{2}$ $H_{1} \cdot \mu_{1}<\mu_{2}$ C. $\mathrm{H}_{0} \mu_{1}=\mu_{2}$ D. $\begin{array}{l}H_{0} \quad \mu_{1} \neq \mu_{2} \\ \\ H_{1} \quad \mu_{1}=\mu_{2}\end{array}$ $\mathrm{H}_{1} \cdot \mu_{1} \neq \mu_{2}$
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Solution

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Solution Steps

Step 1: Hypotheses Formulation

We are testing whether there is a difference in the mean number of partners between the two regions. The hypotheses are formulated as follows:

  • Null Hypothesis (\(H_0\)): \(\mu_1 = \mu_2\) (There is no difference in the mean number of partners)
  • Alternative Hypothesis (\(H_1\)): \(\mu_1 \neq \mu_2\) (There is a difference in the mean number of partners)
Step 2: Test Statistic Calculation

The test statistic is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]

Where:

  • \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means of the two regions.
  • \(SE\) is the standard error, calculated as:

\[ SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = 0.41 \]

Substituting the values, we find:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{0.41} = 6.85 \]

Step 3: Degrees of Freedom

The degrees of freedom (\(df\)) for the test is calculated as:

\[ df = n_1 + n_2 - 2 = 20 + 20 - 2 = 38 \]

Step 4: P-value Calculation

The p-value is calculated using the formula:

\[ P = 2(1 - T(|t|)) = 2(1 - T(6.85)) = 0.0 \]

Step 5: Critical Value

The critical value for a two-tailed test at a significance level of \(\alpha = 0.05\) with \(df = 38\) is:

\[ \text{Critical Value} = 2.02 \]

Step 6: Conclusion

Since the calculated test statistic \(t = 6.85\) exceeds the critical value \(2.02\) and the p-value \(0.0\) is less than the significance level \(0.05\), we reject the null hypothesis.

Final Answer

The hypotheses are:

  • \(H_0: \mu_1 = \mu_2\)
  • \(H_1: \mu_1 \neq \mu_2\)

The test statistic is: \[ \boxed{t = 6.85} \]

The degrees of freedom is: \[ \boxed{df = 38} \]

The p-value is: \[ \boxed{P = 0.0} \]

The critical value is: \[ \boxed{2.02} \]

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