A tire manufacturer guarantess that its tires will last for at least 60,000 miles. If the tire fails earlier, the manufacturer offers a prorated replacement. Too many such replacements are bad for business, so the manufacturer tests tires at random to see whether corrective action is needed on the production line. A random sample of (n = 49 tires). is examined. The average durability is (xbar = 58,000 miles), with a standard deviation of (s = 7000 miles). At the .05 level should the manufacturer be satisfied that its claim is valid? Be sure to state your hypothesis and your basis for rejecting or not rejecting the null hypothesis (sample Z test statistic and rejection threshold)
Use a one-sample z-test.
Formula:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With the data in your problem:
z = (58000 - 60000)/(7000/√49)
Finish the calculation. If the test statistic exceeds the critical value you find in a z-table, reject the null. If the test statistic does not exceed the critical value, do not reject the null.
I hope this will help get you started.
To determine whether the manufacturer should be satisfied with its claim, we need to conduct a hypothesis test. Here's how we can do it step by step:
Step 1: Formulate the hypotheses:
- Null Hypothesis (H0): The average durability of the tires is 60,000 miles (claim is valid).
- Alternative Hypothesis (H1): The average durability of the tires is less than 60,000 miles (claim is not valid).
Step 2: Determine the significance level (α):
The problem states that the significance level is 0.05, which means we are willing to have a 5% chance of making a Type I error (rejecting the null hypothesis when it is actually true).
Step 3: Calculate the test statistic:
We will use the sample mean, sample standard deviation, sample size, and the null hypothesis to calculate the test statistic.
The formula for calculating the test statistic (Z) for a single sample mean is:
Z = (xbar - μ) / (s / sqrt(n))
Where:
- xbar is the sample mean (58,000 miles)
- μ is the population mean (claim) (60,000 miles)
- s is the sample standard deviation (7,000 miles)
- n is the sample size (49 tires)
Plugging in the values:
Z = (58,000 - 60,000) / (7,000 / sqrt(49))
Step 4: Find the rejection threshold:
Since the alternative hypothesis is one-tailed (less than), we need to find the critical Z-value for a 0.05 significance level (α) in the left tail of the standard normal distribution. This can be done using a Z-table or a statistical calculator.
The critical Z-value for a 0.05 significance level is approximately -1.645.
Step 5: Compare the test statistic with the rejection threshold:
If the test statistic (Z) is less than the rejection threshold (-1.645), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 6: Make a decision and interpret the results:
Compare the calculated test statistic (Z) from Step 3 with the rejection threshold from Step 4.
If Z < -1.645, we reject the null hypothesis.
If Z ≥ -1.645, we fail to reject the null hypothesis.
Therefore, the decision about whether the manufacturer should be satisfied with its claim will depend on the value of the test statistic (Z) in Step 3, whether it is less than or greater than -1.645.