Math Problem Statement

A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.67. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59. Conduct the following test of hypothesis using the 0.08 significance level.

H0: μ1 ≤ μ2

H1: μ1 > μ2

what is the p value

Solution

To test the hypothesis:

  • H₀: μ₁ ≤ μ₂
  • H₁: μ₁ > μ₂

We will use the two-sample z-test because we know the population standard deviations.

Steps for conducting the test:

  1. Calculate the test statistic: The formula for the z-test for two population means is:

    z=(xˉ1xˉ2)σ12n1+σ22n2z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}

    Where:

    • xˉ1=2.67\bar{x}_1 = 2.67 (sample mean from population 1)
    • xˉ2=2.59\bar{x}_2 = 2.59 (sample mean from population 2)
    • σ1=0.75\sigma_1 = 0.75 (population standard deviation for population 1)
    • σ2=0.66\sigma_2 = 0.66 (population standard deviation for population 2)
    • n1=65n_1 = 65 (sample size for population 1)
    • n2=50n_2 = 50 (sample size for population 2)
  2. Substitute the values into the formula:

    z=(2.672.59)0.75265+0.66250z = \frac{(2.67 - 2.59)}{\sqrt{\frac{0.75^2}{65} + \frac{0.66^2}{50}}}

  3. Calculate the components:

    • The numerator is: 2.672.59=0.082.67 - 2.59 = 0.08
    • The denominator: 0.75265+0.66250=0.562565+0.435650=0.0086538+0.008712=0.0173658\sqrt{\frac{0.75^2}{65} + \frac{0.66^2}{50}} = \sqrt{\frac{0.5625}{65} + \frac{0.4356}{50}} = \sqrt{0.0086538 + 0.008712} = \sqrt{0.0173658} 0.01736580.1317\sqrt{0.0173658} \approx 0.1317
  4. Calculate the z-statistic:

    z=0.080.13170.607z = \frac{0.08}{0.1317} \approx 0.607

  5. Find the p-value: Since this is a one-tailed test (because the alternative hypothesis is H1:μ1>μ2H_1: \mu_1 > \mu_2), we will find the area to the right of z=0.607z = 0.607.

    Using standard z-tables or a z-calculator, the p-value for z=0.607z = 0.607 is approximately 0.271.

Conclusion:

  • p-value ≈ 0.271

Since the p-value (0.271) is greater than the significance level (0.08), we fail to reject the null hypothesis.

Would you like further details or have any questions? Here are 5 related questions for you:

  1. What is the difference between a one-tailed and a two-tailed hypothesis test?
  2. How do you interpret the significance level (α) in hypothesis testing?
  3. What assumptions are made when conducting a z-test for two population means?
  4. How would the conclusion change if the significance level were 0.10 instead of 0.08?
  5. What is the difference between a p-value and the test statistic in hypothesis testing?

Tip: For hypothesis testing, always ensure your data meets the assumptions for the test you're using, especially when dealing with sample sizes and population standard deviations.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Z-Test
Two-Sample Comparison
P-Value
Normal Distribution

Formulas

z = (x̄1 - x̄2) / sqrt((σ1² / n1) + (σ2² / n2))

Theorems

Central Limit Theorem
Z-Test for Two Population Means

Suitable Grade Level

Grades 11-12 (Advanced)