Conduct the stated hypothesis test for  μ 1− μ 2.
Assume that the samples are independent and randomly selected from normal populations with equal population variances ( σ 12= σ 22)
.
H0 :  μ 1− μ 2=0
H1 :  μ 1− μ 2 < 0
α =0.005
n1=27
x̄ 1=8.72
s1=1.37
n2=25
x̄ 2=9.37
s2=1.23
T-Distribution Table
a. Calculate the test statistic.
t=

0.000
Round to three decimal places if necessary
b. Determine the critical value(s) for the hypothesis test.
+
Round to three decimal places if necessary
c. Conclude whether to reject the null hypothesis or not based on the test statistic.
Reject
Fail to Reject

To conduct the hypothesis test for the difference between two means, we'll use the following information and steps:

### Hypotheses
- **Null Hypothesis (H₀)**: μ₁ - μ₂ = 0
- **Alternative Hypothesis (H₁)**: μ₁ - μ₂ < 0

### Given Data
- **Sample 1**: 
  - n₁ = 27
  - x̄₁ = 8.72
  - s₁ = 1.37
- **Sample 2**:
  - n₂ = 25
  - x̄₂ = 9.37
  - s₂ = 1.23
- **Significance Level (α)**: 0.005

### Assumptions
- The samples are independent and randomly selected.
- The populations are normally distributed with equal variances (σ₁² = σ₂²).

#### (a) **Calculate the test statistic**

Since the population variances are assumed to be equal, we can use the pooled variance:

1. **Pooled variance (sp²)**:
   \[ s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} \]

2. **Test statistic (t)**:
   \[ t = \frac{(x̄_1 - x̄_2) - (μ_1 - μ_2)}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]

Let's calculate the values:

1. **Pooled Variance Calculation**:
   \[ s_p^2 = \frac{(27 - 1)(1.37)^2 + (25 - 1)(1.23)^2}{27 + 25 - 2} \]

2. **Test Statistic Calculation**:
   \[ t = \frac{(8.72 - 9.37) - 0}{s_p \sqrt{\frac{1}{27} + \frac{1}{25}}} \]

Calculating the pooled variance and test statistic:

\[ s_p^2 = \frac{(26)(1.8769) + (24)(1.5129)}{50} \]
\[ s_p^2 = \frac{48.7994 + 36.3096}{50} \]
\[ s_p^2 = 1.69982 \]
\[ s_p = \sqrt{1.69982} \]
\[ s_p \approx 1.3038 \]

Now, the test statistic:

\[ t = \frac{(8.72 - 9.37)}{1.3038 \sqrt{\frac{1}{27} + \frac{1}{25}}} \]
\[ t = \frac{-0.65}{1.3038 \sqrt{0.037037 + 0.04}} \]
\[ t = \frac{-0.65}{1.3038 \sqrt{0.077037}} \]
\[ t = \frac{-0.65}{1.3038 \times 0.27755} \]
\[ t = \frac{-0.65}{0.36187} \]
\[ t \approx -1.796 \]

#### (b) **Determine the critical value(s) for the hypothesis test**

For a left-tailed test at α = 0.005 with degrees of freedom df = n₁ + n₂ - 2 = 50, we use the t-distribution table to find the critical value.

Critical value (t₀.005, 50) ≈ -2.678 (obtained from t-distribution table)

#### (c) **Conclusion**

- **Test Statistic (t)**: -1.796
- **Critical Value**: -2.678

Since the test statistic is not less than the critical value (t > -2.678), we **fail to reject the null hypothesis**.

**Conclusion**: There is insufficient evidence at the 0.005 level of significance to conclude that μ₁ - μ₂ is less than zero.

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Do you want more details or have any questions?

Here are 8 related questions:
1. What are the implications of failing to reject the null hypothesis in this context?
2. How would the results change if the alternative hypothesis was two-tailed?
3. What are the assumptions underlying the use of the pooled variance t-test?
4. How do you determine the degrees of freedom for this test?
5. What would happen if the population variances were not equal?
6. Can this test be used if the data is not normally distributed? Why or why not?
7. How does the choice of significance level affect the hypothesis test conclusion?
8. What is the effect size, and how can it be calculated in this scenario?

**Tip:** Always verify the assumptions of your statistical test, such as normality and equal variances, to ensure the validity of your results.

Learn how to conduct a hypothesis test for the difference between two means using given sample data. This detailed explanation covers calculating the test statistic, determining critical values, and interpreting the results. Suitable for advanced undergraduate students studying statistics.

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