I need someone to explain HOW to do this..I don't want just the answer I have got to figure out how to do this before my mid-term, and I'm completely clueless!
3. 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 believes that the tire actually lasts more than 40,000 miles. A sample 49 tires revealed 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.
Answer these questions:
3a) What is the H0? Null hypothesis, hypothesis about a population of the relationship among populations
3b) What is the H1? alternative hypothesis, hypothesis about parameters that is accepted if the null hypothesis is rejected
3c) What is the decision rule?
3d) What is the calculated value of z?
3e) What is your decision?
Ho: mean = 40,000
H1: mean > 40,000
Z = (score-mean)/SEm
SEm = SD/√n
Since you are only concerned with greater than 40,000, you will reject Ho if Z falls in the top 5%.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
I'll let you calculate Z and come to your own decision.
To answer these questions and test the hypothesis, we need to follow a few steps. Let's go through them one by one:
3a) The null hypothesis H0 states that the mean number of miles the rainmaker tire lasts is equal to or less than 40,000 miles.
So, the null hypothesis H0 is: μ ≤ 40,000 miles
3b) The alternative hypothesis H1 states that the mean number of miles the rainmaker tire lasts is greater than 40,000 miles.
So, the alternative hypothesis H1 is: μ > 40,000 miles
3c) The decision rule is determined by selecting an appropriate significance level. In this case, the significance level is given as 0.05. It means that we are willing to accept a 5% chance of making a Type I error, which is the probability of rejecting the null hypothesis when it is actually true.
For a one-tailed test with a significance level of 0.05, the critical value can be found using a z-table. Since we are testing the claim that the mean number of miles is greater than 40,000, we are looking for the z-value that corresponds to the 0.95 percentile. In this case, the critical value is approximately 1.645.
3d) To calculate the value of z, we use the sample mean and standard deviation provided. The formula for calculating z-score is:
z = (sample mean - population mean) / (standard deviation / square root of sample size)
In this case, the sample mean is 45,000 miles, the population mean (specified by the null hypothesis) is 40,000 miles, the standard deviation is 15,000 miles, and the sample size is 49.
z = (45,000 - 40,000) / (15,000 / sqrt(49))
Simplifying the equation, we get:
z = (5,000) / (15,000 / 7)
z = (5,000) / (2,143)
z ≈ 2.33
So, the calculated value of z is approximately 2.33.
3e) To make a decision, we compare the calculated value of z to the critical value. If the calculated value is greater than or equal to the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
In this case, the calculated value of z is 2.33, which is greater than the critical value of 1.645. Therefore, we reject the null hypothesis.
So, in conclusion, we can conclude that there is enough evidence to suggest that the mean number of miles the rainmaker tire lasts is greater than 40,000 miles based on the sample data provided.