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 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. $H_{0} \mu_{1} \geq \mu_{2}$ B. $H_{0} \mu_{1} \leq \mu_{2}$ $H_{1}, \mu_{1}<\mu_{2}$ $H_{1} \cdot \mu_{1}>\mu_{2}$
C. $\mathrm{H}_{0} \mu_{1}=\mu_{2}$ D. $\mathrm{H}_{0} \mu_{1} \neq \mu_{2}$ $H_{1} \mu_{1} \neq \mu_{2}$ $H_{1} \mu_{1}=\mu_{2}$
Solution
Solution Steps
Step 1: Hypotheses
We are testing the following hypotheses:
Null Hypothesis: \( H_0: \mu_1 = \mu_2 \) (There is no difference in the mean number of partners between the two regions)
Alternative Hypothesis: \( H_1: \mu_1 \neq \mu_2 \) (There is a difference in the mean number of partners between the two regions)
Step 2: Calculate Standard Error
The standard error (SE) is calculated as follows:
\[
SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = 0.4552
\]
Step 3: Calculate Test Statistic
The test statistic \( t \) is computed using the formula:
\[
t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = 1.4278
\]
Step 4: Degrees of Freedom
The degrees of freedom \( df \) for the test is given by:
\[
df = n_1 + n_2 - 2 = 20 + 20 - 2 = 38
\]
Step 5: Calculate P-value
The P-value is calculated as:
\[
P = 2(1 - T(|t|)) = 2(1 - T(1.4278)) = 0.1615
\]
Step 6: Critical Value
The critical value for a two-tailed test at the \( \alpha = 0.05 \) significance level with \( df = 38 \) is:
\[
\text{Critical Value} = 2.0244
\]
Step 7: Conclusion
Since the P-value \( 0.1615 \) is greater than the significance level \( \alpha = 0.05 \), we fail to reject the null hypothesis. Therefore, there is no evidence of a difference in the mean number of partners between the two regions.
Final Answer
\(\boxed{\text{Fail to reject } H_0: \text{ There is no evidence of a difference in the mean number of partners between the two regions.}}\)