Math Problem Statement

Validate the hypothesis that product A is superior to product B in terms of performance. A sample of 20 items of product A is having mean life of 12 months with standard deviation 15 days where as product B is having mean life of 10 months with standard deviation 10 days. Use p – value and validate the hypothesis.

Solution

To validate the hypothesis that product A is superior to product B in terms of performance, we can set up a hypothesis test. Since we're comparing the means of two independent samples, we will perform a two-sample t-test. Here’s a step-by-step approach to solving the problem:

Step 1: State the Hypotheses

  • Null Hypothesis (H₀): The mean life of product A is less than or equal to the mean life of product B. H0:μAμBH_0: \mu_A \leq \mu_B

  • Alternative Hypothesis (H₁): The mean life of product A is greater than the mean life of product B. H1:μA>μBH_1: \mu_A > \mu_B

This is a one-tailed test because we are specifically testing if product A is superior (has a greater mean life) than product B.

Step 2: Define the Parameters

  • Sample size for product A (n₁) = 20
  • Mean life of product A (Xˉ1\bar{X}_1) = 12 months
  • Standard deviation for product A (s₁) = 15 days = 15/30 = 0.5 months
  • Sample size for product B (n₂) = 20
  • Mean life of product B (Xˉ2\bar{X}_2) = 10 months
  • Standard deviation for product B (s₂) = 10 days = 10/30 = 0.33 months

Step 3: Calculate the Test Statistic

We will use the formula for the two-sample t-test assuming unequal variances (since the standard deviations are different):

t=(Xˉ1Xˉ2)s12n1+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=(1210)0.5220+0.33220t = \frac{(12 - 10)}{\sqrt{\frac{0.5^2}{20} + \frac{0.33^2}{20}}}

Let me compute that.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually compute the test statistic using the following values:

t=(1210)0.5220+0.33220t = \frac{(12 - 10)}{\sqrt{\frac{0.5^2}{20} + \frac{0.33^2}{20}}}

Once the test statistic is obtained, you can compare it against the critical t-value (using a t-table for 38 degrees of freedom) or calculate the p-value to validate the hypothesis.

Let me know if you want further explanation on how to proceed, or if you have any questions on calculating the test statistic.

Here are 5 related questions for deeper exploration:

  1. What is the difference between a one-tailed and two-tailed hypothesis test?
  2. How is the degree of freedom calculated for a two-sample t-test?
  3. What is the significance level, and how does it relate to the p-value?
  4. Can we use a z-test instead of a t-test for this scenario?
  5. How do you interpret confidence intervals in hypothesis testing?

Tip: When performing hypothesis testing, always ensure the assumptions of the test (e.g., normality, independence) are met for accurate results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
t-Test

Formulas

Two-sample t-test formula: t = (X̄₁ - X̄₂) / sqrt((s₁²/n₁) + (s₂²/n₂))

Theorems

Central Limit Theorem
t-Distribution Theorem

Suitable Grade Level

Undergraduate Statistics