A football coach claims that the average weight of all the opposing teams’ members is 225 pounds. For a test of the claim, a sample of 50 players is taken from all the opposing teams. The mean is found to be 230 pounds, and the standard deviation is 15 pounds. At a = 0.01, test the coach’s claim. Find the P-value and make the decision.
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
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. Is it larger or smaller than .01?
188-9
To test the coach's claim, we can use a hypothesis test. The null hypothesis (H₀) would be that the average weight of all the opposing teams' members is equal to 225 pounds, while the alternative hypothesis (H₁) would be that it is greater than 225 pounds.
The next step is to calculate the test statistic, which in this case is the z-score. The formula for the z-score is:
z = (sample mean - population mean) / (standard deviation / √n)
Plugging in the given values:
z = (230 - 225) / (15 / √50)
z = 5 / (15 / 7.07)
z = 5 / 2.121 = 2.36 (rounded to two decimal places)
Next, we need to find the p-value associated with the test statistic. The p-value represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed test statistic, assuming the null hypothesis is true.
Since the alternative hypothesis is that the average weight is greater than 225 pounds, we will be conducting a one-tailed test. At a significance level of α = 0.01, we need to find the area in the right tail of the standard normal distribution corresponding to a z-score of 2.36.
Using a standard normal distribution table or a calculator, we find that the area to the right of z = 2.36 is approximately 0.0093.
This p-value of 0.0093 is less than the significance level of 0.01 (α), which means it falls in the rejection region. Therefore, we reject the null hypothesis.
In conclusion, there is sufficient evidence to support the coach's claim that the average weight of all the opposing teams' members is greater than 225 pounds.