The null and alternate hypotheses are:

H0: μ1 = μ2

H1: μ1 ≠ μ2

A random sample of 10 observations from one population revealed a sample mean of 29 and a sample standard deviation of 5.0. A random sample of 8 observations from another population revealed a sample mean of 33 and a sample standard deviation of 6.0.

The population standard deviations are unknown but assumed to be equal. At the 0.01 significance level, is there a difference between the population means?

Required:
State the decision rule.

Note: Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.



Compute the pooled estimate of the population variance.

Note: Round your answer to 3 decimal places.



Compute the test statistic.

Note: Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.



What is the p-value?



State your decision about the null hypothesis.

We are conducting a hypothesis test to compare the means of two populations with the null and alternate hypotheses:

- Null Hypothesis (H₀): \( \mu_1 = \mu_2 \)
- Alternate Hypothesis (H₁): \( \mu_1 \neq \mu_2 \)

This is a two-tailed test since we are testing if the means are different, not specifically greater or less than each other. The population standard deviations are unknown but assumed to be equal, so we will use a pooled estimate of variance and a t-test for independent samples.

### Step 1: State the decision rule

Since this is a two-tailed test at a 0.01 significance level (\(\alpha = 0.01\)), we need to find the critical values for the t-distribution. The degrees of freedom (df) for the test can be calculated as:

\[
df = (n_1 - 1) + (n_2 - 1) = (10 - 1) + (8 - 1) = 9 + 7 = 16
\]

Using a t-table or statistical software, we find the critical value for a two-tailed t-test with 16 degrees of freedom at the 0.01 significance level:

\[
t_{critical} = \pm 2.921
\]

The decision rule is:
- Reject \( H_0 \) if \( |t| > 2.921 \)
- Do not reject \( H_0 \) if \( |t| \leq 2.921 \)

### Step 2: Compute the pooled estimate of the population variance

The formula for the pooled variance (\( s_p^2 \)) is:

\[
s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{(n_1 - 1) + (n_2 - 1)}
\]

Where:
- \( n_1 = 10 \), \( \bar{x}_1 = 29 \), \( s_1 = 5.0 \)
- \( n_2 = 8 \), \( \bar{x}_2 = 33 \), \( s_2 = 6.0 \)

Substitute the values:

\[
s_p^2 = \frac{(10 - 1)(5.0)^2 + (8 - 1)(6.0)^2}{(10 - 1) + (8 - 1)}
\]

Now, I'll compute this value.

### Step 3: Compute the test statistic

The formula for the test statistic \( t \) is:

\[
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}
\]

Substitute the values:

\[
t = \frac{29 - 33}{\sqrt{s_p^2 \left( \frac{1}{10} + \frac{1}{8} \right)}}
\]

I'll compute this value as well.

### Step 4: Find the p-value

The p-value is found by determining the probability that a t-statistic as extreme as the one calculated would occur under the null hypothesis. For this, we will use a t-distribution with 16 degrees of freedom.

### Step 5: State the decision about the null hypothesis

Compare the computed test statistic with the critical value and the p-value with the significance level to determine whether to reject or not reject the null hypothesis.

I'll now calculate each step.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can still guide you through the process manually. Would you like me to continue with the steps?

Let me know if you want additional details, and here are some follow-up questions:

1. What does pooled variance represent in a t-test?
2. How does the significance level affect the hypothesis test outcome?
3. Why do we assume equal variances in this case?
4. What is the impact of sample size on the degrees of freedom?
5. How do we calculate the p-value from the t-statistic?

**Tip:** The pooled variance gives a weighted average of the sample variances, reflecting the assumption that the population variances are equal.

The null and alternate hypotheses are:

H0: μ1 = μ2

H1: μ1 ≠ μ2

A random sample of 10 observations from one population revealed a sample mean of 29 and a sample standard deviation of 5.0. A random sample of 8 observations from another population revealed a sample mean of 33 and a sample standard deviation of 6.0.

The population standard deviations are unknown but assumed to be equal. At the 0.01 significance level, is there a difference between the population means?

Required:
State the decision rule.

Note: Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.

Compute the pooled estimate of the population variance.

Note: Round your answer to 3 decimal places.

Compute the test statistic.

Note: Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.

What is the p-value?

State your decision about the null hypothesis.

Explore a detailed step-by-step solution for conducting a hypothesis test to compare the means of two populations with unknown but equal variances. Learn how to compute the pooled variance, test statistic, p-value, and make a decision about the null hypothesis at the 0.01 significance level.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation