Math Problem Statement

A manufacturer claims that the average lifetime of their new brand of lightbulbs is 1,500 hours. However, a consumer protection agency suspects this claim might be exaggerated and decides to test the manufacturer’s claim. They take a random sample of 50 lightbulbs and find the sample mean lifetime to be 1,480 hours, with a known population standard deviation of 60 hours. Using a 5% significance level, test whether the true mean lifetime differs from 1,500 hours. Answer the following i) State the null hypothesis and the alternative hypothesis for the two-sided mean test. ii) What appropriate test statistic will be suitable for the hypothesis test? Why? C) Calculate the test statistic. iii) Determine the critical Z-value at the 5% significance level iv) State whether to reject or fail to reject the null hypothesis and provide conclusion based on the problem.

Solution

To address this hypothesis testing question, we’ll go through each part systematically.

Given Information:

  • Claimed mean lifetime of lightbulbs, μ0=1500\mu_0 = 1500 hours.
  • Sample mean lifetime, xˉ=1480\bar{x} = 1480 hours.
  • Population standard deviation, σ=60\sigma = 60 hours.
  • Sample size, n=50n = 50.
  • Significance level, α=0.05\alpha = 0.05.

Part (i): State the Null Hypothesis and Alternative Hypothesis

For a two-sided hypothesis test, we want to determine if the true mean lifetime differs from 1,500 hours.

  • Null Hypothesis H0H_0: μ=1500\mu = 1500
  • Alternative Hypothesis HaH_a: μ1500\mu \neq 1500

Part (ii): Test Statistic

Since the population standard deviation σ\sigma is known and the sample size is relatively large (n=50n = 50), we will use the Z-test for testing the hypothesis.

The test statistic for a Z-test is calculated as: Z=xˉμ0σ/nZ = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}

This test is appropriate because:

  1. The population standard deviation is known.
  2. The sample size is large (n30n \geq 30), so we can assume the sampling distribution of the sample mean is approximately normal.

Part (iii): Calculate the Test Statistic

Substituting the values into the formula: Z=1480150060/50=2060/7.071=208.4852.36Z = \frac{1480 - 1500}{60 / \sqrt{50}} = \frac{-20}{60 / 7.071} = \frac{-20}{8.485} \approx -2.36

Part (iv): Determine the Critical Z-Value at the 5% Significance Level

For a two-tailed test at a 5% significance level, the critical Z-values are ±1.96\pm 1.96.

Part (v): Conclusion

To decide whether to reject the null hypothesis, compare the calculated Z-value with the critical Z-value.

  • Calculated Z-value: 2.36-2.36
  • Critical Z-value: ±1.96\pm 1.96

Since 2.36-2.36 falls outside the range [1.96,1.96][-1.96, 1.96], we reject the null hypothesis.

Conclusion

At a 5% significance level, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the true mean lifetime of the lightbulbs differs from the manufacturer’s claim of 1,500 hours.


Would you like further clarification on any of these steps, or need additional details?

Here are 5 related questions to expand on this topic:

  1. What would change if the sample size was smaller, say 20, instead of 50?
  2. How would the conclusion differ if the significance level was changed to 1%?
  3. What assumptions are needed to ensure the validity of this Z-test?
  4. How would the test change if the population standard deviation were unknown?
  5. What if we wanted to test if the true mean lifetime is less than 1,500 hours instead?

Tip: When performing hypothesis tests, always check if the population standard deviation is known, as this determines whether to use a Z-test or a t-test.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Z-Test
Confidence Intervals
Significance Levels

Formulas

Z = (x̄ - μ₀) / (σ / √n)

Theorems

Central Limit Theorem
Z-Test for Hypothesis Testing

Suitable Grade Level

College/University Level (Introductory Statistics)