Math Problem Statement

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim?

Solution

To assess whether the data is inconsistent with the tire manufacturer’s claim (that the average tire lifespan is at least 50,000 miles), we can perform a hypothesis test.

Step 1: Set up hypotheses

The hypotheses are:

  • Null Hypothesis (H0)(H_0): μ50,000\mu \geq 50,000
  • Alternative Hypothesis (Ha)(H_a): μ<50,000\mu < 50,000

This is a left-tailed test, as we are testing if the mean lifespan is significantly less than 50,000 miles.

Step 2: Identify known values

  • Population standard deviation σ=8,000\sigma = 8,000
  • Sample size n=28n = 28
  • Sample mean xˉ=46,500\bar{x} = 46,500
  • Population mean claimed by the manufacturer μ0=50,000\mu_0 = 50,000
  • Significance level α=0.05\alpha = 0.05

Since the population standard deviation is known, we can use a z-test.

Step 3: Calculate the test statistic

The formula for the z-test statistic is: z=xˉμ0σnz = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}

Substitute the values: z=46,50050,0008,00028z = \frac{46,500 - 50,000}{\frac{8,000}{\sqrt{28}}}

  1. Calculate the standard error (SE): SE=8,000281,511.86SE = \frac{8,000}{\sqrt{28}} \approx 1,511.86

  2. Calculate the z-value: z=46,50050,0001,511.863,5001,511.862.31z = \frac{46,500 - 50,000}{1,511.86} \approx \frac{-3,500}{1,511.86} \approx -2.31

Step 4: Determine the critical value

For a significance level of α=0.05\alpha = 0.05 in a one-tailed test, the critical value for zz is approximately 1.645-1.645.

Step 5: Make a decision

  • If z<1.645z < -1.645, we reject the null hypothesis.
  • Calculated z-value = 2.31-2.31, which is less than 1.645-1.645.

Conclusion

Since z=2.31z = -2.31 is less than 1.645-1.645, we reject the null hypothesis. This indicates that the data is highly inconsistent with the manufacturer's claim. There is sufficient evidence at the 0.05 significance level to conclude that the average lifespan of the tires is likely less than 50,000 miles.

Would you like further details on any part of this solution?


  1. What is a z-test and when is it used in hypothesis testing?
  2. Why do we use the population standard deviation in this test instead of the sample standard deviation?
  3. What does it mean if the test statistic is less than the critical value?
  4. How does the sample size affect the outcome of the hypothesis test?
  5. What would change in the solution if the significance level were 0.01 instead of 0.05?

Tip: In hypothesis testing, carefully distinguish between population and sample standard deviations—using the wrong one can lead to incorrect conclusions.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
Z-Test

Formulas

z = (x̄ - μ0) / (σ / √n)

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 11-12