one of the major U.S. tire makers wishes to review its warranty for their rainmaker tire. The warranty is for 40,000 miles. The tire company beleives that the tire actually lasts more than 40,000 miles. A sample of 49 tires reveales that the mean number of miles is 45,000 miles with a standard deviation of 15,000 miles. Test the hypothesis with a 0.05 significance level. What is the decision rule? What is the calculated value of z? what is our decision?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)(but you can use n)
If only one SD is provided, you can use just that to determine SEdiff.
You can do the calculations in thousands, leaving off the extra three zeros.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
Ho: mean = 40,000
Ha: mean > 40,000
The decision is yours.
To test the hypothesis, we need to set up the null and alternative hypotheses:
Null hypothesis (H0): The mean number of miles the tire lasts is 40,000 miles.
Alternative hypothesis (Ha): The mean number of miles the tire lasts is greater than 40,000 miles.
Next, we determine the significance level, which is given as 0.05.
The decision rule for a one-tailed test with a significance level of 0.05 is as follows:
- If the calculated value of the test statistic is greater than the critical value (z > z-critical), we reject the null hypothesis.
- If the calculated value of the test statistic is less than or equal to the critical value (z ≤ z-critical), we fail to reject the null hypothesis.
To calculate the test statistic (z), we can use the formula:
z = (sample mean - population mean) / (standard deviation / sqrt(sample size))
Given:
Sample mean (x̄) = 45,000 miles
Population mean (µ) = 40,000 miles
Standard deviation (σ) = 15,000 miles
Sample size (n) = 49
Calculating z:
z = (45000 - 40000) / (15000 / sqrt(49))
z = 5000 / (15000 / 7)
z ≈ 1.87
To find the critical value, we look up the z-value corresponding to a significance level of 0.05 for a one-tailed test. In this case, the critical value (z-critical) is approximately 1.645.
Since the calculated value of z (1.87) is greater than the critical value (1.645), we reject the null hypothesis.
Therefore, our decision is that there is sufficient evidence to support the claim that the mean number of miles the tire lasts is greater than 40,000 miles.
To test the hypothesis, we need to follow a few steps:
Step 1: Formulate the null and alternative hypotheses:
- Null hypothesis (H₀): The mean number of miles for the Rainmaker tire is 40,000 miles.
- Alternative hypothesis (H₁): The mean number of miles for the Rainmaker tire is greater than 40,000 miles.
Step 2: Determine the significance level:
In this case, the significance level is given as 0.05.
Step 3: Determine the test statistic:
Since we know the population standard deviation (15,000 miles) and the sample size (49), we can use the z-test.
Step 4: Set the decision rule:
The decision rule depends on the type of test (one-tailed or two-tailed) and the significance level. In this case, since the alternative hypothesis is that the mean is greater than 40,000 miles, we have a one-tailed test with a right tail.
At a significance level of 0.05, the critical z-value (obtained from the z-table or calculator) is approximately 1.645.
Step 5: Calculate the test statistic (z-value):
The formula to calculate the z-value is:
z = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
Given:
Sample mean (x̄) = 45,000 miles
Population mean (μ) = 40,000 miles
Population standard deviation (σ) = 15,000 miles
Sample size (n) = 49
Calculating the z-value:
z = (45,000 - 40,000) / (15,000 / sqrt(49))
z = 5,000 / (15,000 / 7)
z ≈ 2.333
Step 6: Make a decision:
If the calculated z-value falls within the critical region, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated z-value is approximately 2.333, which is greater than the critical z-value of 1.645. Therefore, the calculated z-value falls within the critical region.
Decision: We reject the null hypothesis (H₀) in favor of the alternative hypothesis (H₁). The evidence suggests that the mean number of miles for the Rainmaker tire is greater than 40,000 miles at the 0.05 significance level.