To perform a hypothesis test for the equality of two population means based on the given data, I will follow these steps:

Step 1: Set up the hypotheses

• Null Hypothesis ($H_0$): The population means are equal, $\mu_1 = \mu_2$.
• Alternative Hypothesis ($H_1$): The population means are not equal, $\mu_1 \neq \mu_2$.

This is a two-tailed test.

Step 2: Gather the necessary information

From the updated parameters you gave:

• $n_1 = 214$ (Sample size of Company 1)
• $n_2 = 375$ (Sample size of Company 2)
• $\bar{x}_1 = 3.51$ (Sample mean of Company 1)
• $\bar{x}_2 = 3.24$ (Sample mean of Company 2)
• $S_1 = 0.51$ (Sample standard deviation of Company 1)
• $S_2 = 0.52$ (Sample standard deviation of Company 2)
• Significance level $\alpha = 0.05$

Step 3: Compute the test statistic

The formula for the test statistic when comparing two population means is:

$t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}$

Substituting the values:

$t = \frac{(3.51 - 3.24)}{\sqrt{\frac{0.51^2}{214} + \frac{0.52^2}{375}}}$

Let me calculate this step:

$t = \frac{0.27}{\sqrt{\frac{0.2601}{214} + \frac{0.2704}{375}}}$

$t = \frac{0.27}{\sqrt{0.001215 + 0.000721}}$

$t = \frac{0.27}{\sqrt{0.001936}}$

$t = \frac{0.27}{0.044}$

$t \approx 6.14$

Step 4: Determine the degrees of freedom

We can approximate the degrees of freedom using the following formula:

$df \approx \frac{\left(\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}\right)^2}{\frac{\left(\frac{S_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{S_2^2}{n_2}\right)^2}{n_2 - 1}}$

Let me calculate the degrees of freedom:

$df = \frac{\left(0.001215 + 0.000721\right)^2}{\frac{(0.001215)^2}{214 - 1} + \frac{(0.000721)^2}{375 - 1}}$

$df = \frac{(0.001936)^2}{\frac{0.000001477}{213} + \frac{0.00000052}{374}}$

$df = \frac{0.000003747}{0.000000006935 + 0.00000000139}$

$df \approx 405.8 \approx 406$

Step 5: Find the critical value

For a two-tailed test at $\alpha = 0.05$ and $df = 406$, the critical $t$-value from the $t$-distribution table is approximately $t_{crit} = 1.966$.

Step 6: Compare the test statistic to the critical value

Since the computed test statistic $t \approx 6.14$ is much larger than the critical value $1.966$, we reject the null hypothesis.

Conclusion:

At the 5% significance level, we reject the null hypothesis and conclude that there is sufficient evidence to suggest a significant difference in the customer satisfaction levels between the two companies.

Would you like more details on any step? Here are 5 related questions:

1. How is the degrees of freedom calculation related to the variances in this test?
2. What if the variances were assumed to be equal—how would the test change?
3. Could we perform this test with a different significance level, such as 1%?
4. What are the assumptions underlying this hypothesis test?
5. How does the sample size affect the power of this test?

Tip: When conducting hypothesis tests for two means, always check if the population variances are assumed to be equal or unequal, as it impacts the type of test and formula used.