Math Problem Statement

Test whether μ₁ < μ₂ at the α = 0.02 level of significance for the sample data provided. Population 1: n = 33, x̄ = 103.5, s = 12.3. Population 2: n = 25, x̄ = 114.5, s = 13.2.

Solution

The question asks to test whether μ1<μ2\mu_1 < \mu_2 at the α=0.02\alpha = 0.02 level of significance using the provided sample dataThe question asks to test whether μ1<μ2\mu_1 < \mu_2 at the α=0.02\alpha = 0.02 level of significance using the provided sample

Data Summary:

  • Population 1:

    • n1=33n_1 = 33
    • xˉ1=103.5\bar{x}_1 = 103.5
    • s1=12.3s_1 = 12.3
  • Population 2:

    • n2=25n_2 = 25
    • xˉ2=114.5\bar{x}_2 = 114.5
    • s2=13.2s_2 = 13.2

Hypotheses:

  • H0:μ1=μ2H_0: \mu_1 = \mu_2 (Null Hypothesis)
  • H1:μ1μ2H_1: \mu_1 \neq \mu_2 (Alternative Hypothesis)

This is a two-tailed test because the alternative hypothesis is a non-equality (μ1μ2\mu_1 \neq \mu_2).

Step 1: Calculate the Test Statistic

The formula for the test statistic tt is:

t=xˉ1xˉ2s12n1+s22n2t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}

Substituting the values:

t=103.5114.512.3233+13.2225t = \frac{103.5 - 114.5}{\sqrt{\frac{12.3^2}{33} + \frac{13.2^2}{25}}}

t=11151.2933+174.2425t = \frac{-11}{\sqrt{\frac{151.29}{33} + \frac{174.24}{25}}}

t=114.582+6.97=1111.552=113.43.235t = \frac{-11}{\sqrt{4.582 + 6.97}} = \frac{-11}{\sqrt{11.552}} = \frac{-11}{3.4} \approx -3.235

Step 2: Degrees of Freedom (df)

The degrees of freedom for unequal sample sizes and variances can be approximated using the following formula:

df=(s12n1+s22n2)2(s12n1)2n11+(s22n2)2n21df = \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}}

Let me calculate this.The test statistic tt is approximately 3.236-3.236, and the degrees of freedom (df) is about 49.8.

Step 3: Determine the P-value

For a two-tailed test with a tt-value of 3.236-3.236 and df50df \approx 50, we can find the P-value using a t-distribution table or software.

Let's calculate the P-value.The P-value for this hypothesis test is approximately 0.002.

Step 4: Conclusion

Since the P-value (0.002) is less than the significance level α=0.02\alpha = 0.02, we reject the null hypothesis. This means that there is sufficient evidence at the 0.02 level of significance to conclude that μ1μ2\mu_1 \neq \mu_2.

Would you like more details on any part of this calculation?

5 Follow-up Questions:

  1. How would the test change if the significance level were α=0.05\alpha = 0.05?
  2. Can you explain how the degrees of freedom formula is derived?
  3. What is the impact of sample size on the test statistic?
  4. How do we interpret the P-value in a hypothesis test?
  5. What is the difference between a one-tailed and two-tailed test?

Tip: In hypothesis testing, always compare the P-value with the significance level (α\alpha) to make your decision about rejecting or not rejecting the null hypothesis.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Two-Sample t-Test
P-value Calculation

Formulas

t = (x̄₁ - x̄₂) / sqrt((s₁² / n₁) + (s₂² / n₂))
Degrees of Freedom Formula for Unequal Sample Sizes

Theorems

Central Limit Theorem
t-Distribution

Suitable Grade Level

Undergraduate Level (Statistics)