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, hours.
- Sample mean lifetime, hours.
- Population standard deviation, hours.
- Sample size, .
- Significance level, .
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 :
- Alternative Hypothesis :
Part (ii): Test Statistic
Since the population standard deviation is known and the sample size is relatively large (), we will use the Z-test for testing the hypothesis.
The test statistic for a Z-test is calculated as:
This test is appropriate because:
- The population standard deviation is known.
- The sample size is large (), 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:
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 .
Part (v): Conclusion
To decide whether to reject the null hypothesis, compare the calculated Z-value with the critical Z-value.
- Calculated Z-value:
- Critical Z-value:
Since falls outside the range , 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:
- What would change if the sample size was smaller, say 20, instead of 50?
- How would the conclusion differ if the significance level was changed to 1%?
- What assumptions are needed to ensure the validity of this Z-test?
- How would the test change if the population standard deviation were unknown?
- 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.
Ask a new question for Free
By Image
Drop file here or Click Here to upload
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)
Related Recommendation
Hypothesis Testing for Mean Life of LEDs Using Z-test at 0.05 Significance Level
Hypothesis Test for Population Mean: Z-Test with 0.01 Significance Level
Hypothesis Testing for Light Bulb Life Claim with 0.08 Significance
Hypothesis Testing for Population Mean Lifetime of Light Bulbs
Probability of Sample Mean for Bulb Lifespan in Normal Distribution