You wish to test the following claim (Ha) at a significance level of α=0.005.

      Ho:μ1=μ2
      Ha:μ1>μ2

You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain a sample of size n1=11 with a mean of M1=80.5 and a standard deviation of SD1=10.4 from the first population. You obtain a sample of size n2=28 with a mean of M2=76.1 and a standard deviation of SD2=19.9 from the second population.

What is the p-value for this test? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to four decimal places.)
p-value =

To calculate the \( p \)-value for this two-sample \( t \)-test, where you are testing the claim \( H_a: \mu_1 > \mu_2 \), we need to follow these steps:

### Step 1: Define the hypotheses

- Null hypothesis (\( H_0 \)): \( \mu_1 = \mu_2 \)
- Alternative hypothesis (\( H_a \)): \( \mu_1 > \mu_2 \)

This is a **one-tailed test**.

### Step 2: Calculate the test statistic

Since the sample sizes are small, the standard deviations are not known, and the variances are assumed to be equal, we use the pooled standard deviation for the test statistic.

#### a) Pooled standard deviation

The formula for the pooled standard deviation \( S_p \) is:

\[
S_p = \sqrt{\frac{(n_1 - 1) \cdot SD_1^2 + (n_2 - 1) \cdot SD_2^2}{n_1 + n_2 - 2}}
\]

Where:
- \( n_1 = 11 \), \( n_2 = 28 \)
- \( SD_1 = 10.4 \), \( SD_2 = 19.9 \)

Plugging in the values:

\[
S_p = \sqrt{\frac{(11 - 1) \cdot 10.4^2 + (28 - 1) \cdot 19.9^2}{11 + 28 - 2}}
\]

\[
S_p = \sqrt{\frac{10 \cdot 108.16 + 27 \cdot 396.01}{37}}
\]

\[
S_p = \sqrt{\frac{1081.6 + 10692.27}{37}} = \sqrt{\frac{11773.87}{37}} = \sqrt{318.21} \approx 17.84
\]

#### b) Test statistic \( t \)

Now calculate the test statistic \( t \):

\[
t = \frac{M_1 - M_2}{S_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\]

Where:
- \( M_1 = 80.5 \), \( M_2 = 76.1 \)
- \( n_1 = 11 \), \( n_2 = 28 \)
- \( S_p = 17.84 \)

Plugging in the values:

\[
t = \frac{80.5 - 76.1}{17.84 \cdot \sqrt{\frac{1}{11} + \frac{1}{28}}}
\]

\[
t = \frac{4.4}{17.84 \cdot \sqrt{0.0909 + 0.0357}} = \frac{4.4}{17.84 \cdot \sqrt{0.1266}} = \frac{4.4}{17.84 \cdot 0.3558} = \frac{4.4}{6.35} \approx 0.6929
\]

### Step 3: Degrees of freedom (conservative estimate)

Since you need to use a conservative estimate for the degrees of freedom, we'll use the smaller of \( n_1 - 1 \) or \( n_2 - 1 \). In this case:

\[
df = \min(n_1 - 1, n_2 - 1) = \min(11 - 1, 28 - 1) = \min(10, 27) = 10
\]

### Step 4: Calculate the \( p \)-value

Using a \( t \)-distribution table or a calculator, look up the one-tailed \( p \)-value for \( t = 0.6929 \) with \( df = 10 \).

From the \( t \)-distribution table, for \( t = 0.6929 \) and \( df = 10 \), the \( p \)-value is approximately **0.2510**.

### Step 5: Conclusion

The \( p \)-value for this test is \( \boxed{0.2510} \).

Since the \( p \)-value (0.2510) is much greater than the significance level \( \alpha = 0.005 \), we **fail to reject the null hypothesis**. There is not enough evidence to support the claim that \( \mu_1 > \mu_2 \).

Would you like further details or have any questions?

---

Here are 5 related questions to deepen your understanding:
1. What would change in the calculation if the sample sizes were larger?
2. How would a two-tailed test modify the hypothesis and \( p \)-value interpretation?
3. Can you explain why we use a pooled standard deviation in this test?
4. How would the test change if the standard deviations were known?
5. How do we calculate degrees of freedom for unequal variances?

**Tip**: Always ensure the assumptions of your test match the data (normality, variance equality, etc.) before applying statistical methods.

Solve a two-sample t-test problem where μ1 > μ2 at a significance level of α=0.005. This example involves calculating the pooled standard deviation, test statistic, and p-value using a conservative estimate for degrees of freedom. Sample data includes n1=11, M1=80.5, SD1=10.4, n2=28, M2=76.1, SD2=19.9. Suitable for undergraduate statistics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation