Conduct the stated hypothesis test for  μ 1− μ 2. μ 1− μ 2. Assume that the samples are independent and randomly selected from normal populations with **equal population variances** ( σ 12= σ 22)( σ 12= σ 22).

H0 :  μ 1− μ 2=0H0 :  μ 1− μ 2=0

H1 :  μ 1− μ 2 < 0H1 :  μ 1− μ 2 < 0

α =0.005 α =0.005

n1=29n1=29

x̄ 1=8.51 x̄ 1=8.51

s1=1.26s1=1.26

n2=23n2=23

x̄ 2=9.36 x̄ 2=9.36

s2=1.35s2=1.35

T-Distribution Table

**a.** Calculate the test statistic.

t=t=

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

We are conducting a hypothesis test to compare the difference between two population means, \(\mu_1\) and \(\mu_2\), with the following information:

### Hypotheses:
- Null hypothesis (\(H_0\)): \(\mu_1 - \mu_2 = 0\) (no difference in population means)
- Alternative hypothesis (\(H_1\)): \(\mu_1 - \mu_2 < 0\) (mean of population 1 is less than mean of population 2)

### Given Information:
- Significance level: \(\alpha = 0.005\)
- Sample 1: \(n_1 = 29\), \(\bar{x}_1 = 8.51\), \(s_1 = 1.26\)
- Sample 2: \(n_2 = 23\), \(\bar{x}_2 = 9.36\), \(s_2 = 1.35\)
- Population variances are assumed equal: \(\sigma_1^2 = \sigma_2^2\)

### Steps to solve the problem:

#### a. **Calculate the test statistic.**

Since the population variances are assumed to be equal, we will use the **pooled variance** and a **two-sample t-test** formula. The test statistic \(t\) is calculated as:

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

Where:
- \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means,
- \(n_1\) and \(n_2\) are the sample sizes,
- \(s_p\) is the pooled standard deviation, calculated as:

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

Let’s compute the test statistic.

#### b. **Determine the critical value.**

To determine the critical value for a one-tailed t-test at a significance level \(\alpha = 0.005\), we need the degrees of freedom (df) and the t-distribution table.

\[
\text{df} = n_1 + n_2 - 2
\]

#### c. **Conclude whether to reject the null hypothesis.**

We will compare the test statistic calculated in part (a) to the critical value from part (b). If the test statistic is less than the critical value (in the negative direction), we will reject the null hypothesis. Otherwise, we fail to reject it.

Let's compute the values step by step.### Results:

a. **Test Statistic:**
\[
t = -2.341
\]

b. **Critical Value:**
For a one-tailed test with \(\alpha = 0.005\) and 50 degrees of freedom:
\[
t_{\text{critical}} = -2.678
\]

c. **Conclusion:**
Since the test statistic (\(-2.341\)) is **greater** than the critical value (\(-2.678\)), we **fail to reject** the null hypothesis. This means there is not enough evidence at the \(\alpha = 0.005\) level to conclude that \(\mu_1 < \mu_2\).

Would you like further details or explanations?

---

Here are 5 related questions for deeper exploration:

1. How would the result change if the significance level \(\alpha\) were 0.01?
2. How do we perform a two-tailed test in this scenario?
3. What is the role of the pooled variance in hypothesis testing?
4. How would unequal population variances affect the test?
5. How can the sample sizes influence the test statistic and outcome?

**Tip:** The degrees of freedom in a t-test affect the shape of the t-distribution and hence the critical value. Always check if it's correct for your sample sizes.

Conduct the stated hypothesis test for μ1 − μ2. Assume that the samples are independent and randomly selected from normal populations with equal population variances (σ12 = σ22). H0 : μ1 − μ2 = 0 H1 : μ1 − μ2 < 0 α = 0.005 n1 = 29 x̄1 = 8.51 s1 = 1.26 n2 = 23 x̄2 = 9.36 s2 = 1.35 Calculate the test statistic, determine the critical value, and conclude whether to reject the null hypothesis.

This page explains how to conduct a hypothesis test for the difference between two population means using a two-sample t-test with equal variances. Given α = 0.005, n1 = 29, n2 = 23, x̄1 = 8.51, x̄2 = 9.36, s1 = 1.26, and s2 = 1.35, it calculates the test statistic and critical value, and concludes whether to reject the null hypothesis.

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