Provided below are summary statistics for independent simple random samples from two populations. Use the nonpooled​ t-test and the nonpooled​ t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.
x overbar 1equals13​, s 1equals2​, n 1equals10​, x overbar 2equals14​, s 2equals3​, n 2equals10
a.​ Two-tailed test, alphaequals0.05      b. 95​% confidence interval
Question content area bottom
Part 1
a. What are the hypotheses for the​ t-test?
A.
H Subscript 0​: mu Subscript 1equalsmu Subscript 2
H Subscript a​: mu Subscript 1greater thanmu Subscript 2
B.
H Subscript 0​: mu Subscript 1equalsmu Subscript 2
H Subscript a​: mu Subscript 1not equalsmu Subscript 2
Your answer is correct.C.
H Subscript 0​: mu Subscript 1equalsmu Subscript 2
H Subscript a​: mu Subscript 1less thanmu Subscript 2
D.
H Subscript 0​: mu Subscript 1greater than or equalsmu Subscript 2
H Subscript a​: mu Subscript 1less thanmu Subscript 2
Part 2
Find the test statistic.
tequals

negative 0.877 ​(Round to three decimal places as​ needed.)
Part 3
Find the​ P-value.
Pequals

0.394 ​(Round to four decimal places as​ needed.)
Part 4
What is the conclusion of the hypothesis​ test?
A.
Reject H Subscript 0. There is insufficient evidence that the two means are different.
B.
Reject H Subscript 0. There is sufficient evidence that the two means are different.
C.
Do not reject H Subscript 0. There is sufficient evidence that the two means are different.
D.
Do not reject H Subscript 0. There is insufficient evidence that the two means are different.
Your answer is correct.
Part 5
b. The 95​% confidence interval is from

enter your response here to

enter your response here.
​(Round to three decimal places as​ needed.)

Given the summary statistics and the requirements, let's address each part systematically.

### Part 1: Hypotheses for the t-test
The hypotheses for the two-tailed test are:
- **Null Hypothesis (H₀):** \(\mu_1 = \mu_2\)
- **Alternative Hypothesis (Hₐ):** \(\mu_1 \neq \mu_2\)

So the correct option is:
**B.**
\[ H_0: \mu_1 = \mu_2 \]
\[ H_a: \mu_1 \neq \mu_2 \]

### Part 2: Test Statistic
The formula for the test statistic in a nonpooled t-test is:

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

Plugging in the given values:
- \(\bar{x}_1 = 13\)
- \(\bar{x}_2 = 14\)
- \(s_1 = 2\)
- \(s_2 = 3\)
- \(n_1 = 10\)
- \(n_2 = 10\)

We get:

\[ t = \frac{13 - 14}{\sqrt{\frac{2^2}{10} + \frac{3^2}{10}}} = \frac{-1}{\sqrt{\frac{4}{10} + \frac{9}{10}}} = \frac{-1}{\sqrt{1.3}} = \frac{-1}{1.14} \approx -0.877 \]

### Part 3: P-value
To find the p-value, we need the degrees of freedom, which can be approximated using the following formula for nonpooled t-tests:

\[ \text{df} = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}} \]

Plugging in the values:

\[ \text{df} = \frac{\left( \frac{4}{10} + \frac{9}{10} \right)^2}{\frac{\left( \frac{4}{10} \right)^2}{9} + \frac{\left( \frac{9}{10} \right)^2}{9}} = \frac{1.3^2}{\frac{0.16}{9} + \frac{0.81}{9}} = \frac{1.69}{\frac{0.97}{9}} = \frac{1.69}{0.108} \approx 15.65 \]

Using the t-distribution table or a calculator for \( t = -0.877 \) and \( \text{df} \approx 16 \), the p-value is approximately 0.394.

### Part 4: Conclusion of the Hypothesis Test
Given \( \alpha = 0.05 \) and \( \text{p-value} = 0.394 \):
- Since \( \text{p-value} > \alpha \), we do not reject the null hypothesis.

Thus, the correct conclusion is:
**D.** Do not reject \( H_0 \). There is insufficient evidence that the two means are different.

### Part 5: 95% Confidence Interval
The formula for the confidence interval in a nonpooled t-test is:

\[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, \, \text{df}} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Using \( t_{0.025, 16} \approx 2.12 \) from the t-distribution table:

\[ \text{Margin of Error} = 2.12 \cdot \sqrt{\frac{4}{10} + \frac{9}{10}} = 2.12 \cdot \sqrt{1.3} = 2.12 \cdot 1.14 \approx 2.417 \]

So, the confidence interval is:

\[ (13 - 14) \pm 2.417 = -1 \pm 2.417 \]

Thus, the 95% confidence interval is from:

\[ -3.417 \text{ to } 1.417 \]

### Summary

#### a. Two-tailed test hypotheses:
- **H₀:** \(\mu_1 = \mu_2\)
- **Hₐ:** \(\mu_1 \neq \mu_2\)

#### b. 95% confidence interval:
- From \(-3.417\) to \(1.417\)

Do you want details or have any questions?

### Related Questions
1. How do you interpret the p-value in hypothesis testing?
2. What is the difference between pooled and nonpooled t-tests?
3. Why is it important to check the assumptions of normality and equal variances in t-tests?
4. How does the sample size affect the degrees of freedom in a t-test?
5. What is the impact of outliers on the t-test results?
6. How do you calculate the margin of error for a confidence interval?
7. What are Type I and Type II errors in hypothesis testing?
8. How can you increase the power of a statistical test?

### Tip
Always verify the assumptions of normality and equal variances before applying the t-test to ensure the validity of the results.

Explore a detailed solution using nonpooled t-test and confidence interval methods for hypothesis testing based on summary statistics. Includes step-by-step calculations and interpretations. Suitable for advanced college-level students and statisticians.

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