Independent random samples from approximately normal populations produced the results shown below. Assume that the population variances are unequal. Complete parts a and b.
Full data set   
Sample 1
Sample 2
49
30
32
34
46
64
53
60
55
50
44
22
33
50
53
24
33
34
44
64
Question content area bottom
Part 1
a.
Do the data present sufficient evidence to conclude that mu 1 minus mu 2not equals​0? Use alpha equals 0.02.
Let mu 1 be the mean of population 1 and mu 2 be the mean of population 2. Select the correct hypotheses below.
A.
Upper H 0​: mu 1 minus mu 2equals​0, Upper H Subscript a Baseline : mu 1 minus mu 2 greater than 0
B.
Upper H 0​: mu 1 minus mu 2equals​0, Upper H Subscript a Baseline : mu 1 minus mu 2 not equals 0
Your answer is correct.C.
Upper H 0​: mu 1 minus mu 2equals​0, Upper H Subscript a Baseline : mu 1 minus mu 2 less than 0
Part 2
Now find the test statistic.
tequals
  
enter your response here ​(Round to two decimal places as​ needed.)

To solve this problem, we will follow these steps:

### Part 1: Hypotheses
Given that we need to test whether \(\mu_1 - \mu_2 \neq 0\) at \(\alpha = 0.02\), the correct hypotheses are:

- **Null Hypothesis (H₀)**: \(\mu_1 - \mu_2 = 0\) (There is no difference between the population means)
- **Alternative Hypothesis (Hₐ)**: \(\mu_1 - \mu_2 \neq 0\) (There is a significant difference between the population means)

Therefore, the correct choice is:

\[
\boxed{B. \ H_0: \mu_1 - \mu_2 = 0 \ , \ H_a: \mu_1 - \mu_2 \neq 0}
\]

### Part 2: Test Statistic

We are performing a two-sample \( t \)-test for unequal variances (also known as Welch’s t-test). To compute the test statistic \( t \), we use the formula:

\[
t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

Where:
- \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means,
- \(s_1^2\) and \(s_2^2\) are the sample variances,
- \(n_1\) and \(n_2\) are the sample sizes.

Let's calculate the necessary values:

#### Step 1: Calculate the sample means:
For Sample 1: \(49, 32, 46, 53, 55, 44, 33, 53, 33, 44\)

\[
\bar{X}_1 = \frac{49 + 32 + 46 + 53 + 55 + 44 + 33 + 53 + 33 + 44}{10} = \frac{442}{10} = 44.2
\]

For Sample 2: \(30, 34, 64, 60, 50, 22, 50, 24, 34, 64\)

\[
\bar{X}_2 = \frac{30 + 34 + 64 + 60 + 50 + 22 + 50 + 24 + 34 + 64}{10} = \frac{432}{10} = 43.2
\]

#### Step 2: Calculate the sample variances:
For Sample 1:
\[
s_1^2 = \frac{\sum{(X_1 - \bar{X}_1)^2}}{n_1 - 1}
\]

For Sample 2:
\[
s_2^2 = \frac{\sum{(X_2 - \bar{X}_2)^2}}{n_2 - 1}
\]

Let me calculate the variances and then the test statistic.It seems like I can’t do more advanced data analysis right now. Please try again later. If you want, you can try calculating the variances and test statistic on your own based on the steps I provided.

If you need further help, feel free to ask! Meanwhile, here are some related questions to deepen your understanding:

1. What assumptions are needed for Welch’s t-test?
2. How do you interpret the p-value in hypothesis testing?
3. What’s the difference between a one-tailed and two-tailed test?
4. How does the unequal variance assumption affect the t-test?
5. Why do we use sample variances instead of population variances in hypothesis tests?

**Tip**: Always check the assumptions of normality and variance before selecting the appropriate test method.

Independent random samples from approximately normal populations produced the results shown below. Assume that the population variances are unequal. Complete parts a and b. Part 1: Do the data present sufficient evidence to conclude that μ1 - μ2 ≠ 0? Use α = 0.02. Let μ1 be the mean of population 1 and μ2 be the mean of population 2.

Solve a hypothesis testing problem using the two-sample t-test for unequal variances (Welch’s t-test). The data involves independent random samples, and the objective is to test if the difference in population means (μ1 - μ2) is significant at α = 0.02. Detailed steps for calculating sample means, variances, and test statistics are provided.

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