Answer :
To calculate the P-value limits based on the z-statistic of 2.899 in a one-tailed test, we use the normal distribution table. The lower limit for the P-value is 0.0019, which corresponds to the given z-statistic, and the upper limit is 0.9981, which represents the probability of finding a value greater than the z-score under the normal curve.
The question involves conducting a hypothesis test on the mean high school football score per game in Ohio based on a z-statistic given as 2.899. Since the z-statistic is 2.899 standard deviations above the mean, we are dealing with a one-tailed test (because the hypothesis is that Ohio scores higher, not just different).
To determine the lower and upper limit for the P-value, we need to refer to the standard normal distribution table. A z-score of 2.899 corresponds to a P-value that is the area under the normal curve to the right of the z-score. Unfortunately, standard z-tables typically only give the area to the left. However, since the normal distribution is symmetrical, finding the area to the left for a z-score of -2.899 gives the same numerical value for the area to the right of +2.899.
Consulting a z-table or using statistical software, we can find that the area to the left of -2.899 is approximately 0.0019. Therefore, the P-value for our one-tailed test is the area to the right, which is 1 - 0.0019, or 0.9981. The lower limit for the P-value is the P-value itself (since it cannot be lower), which is 0.0019, and the upper limit is the chance of an observation falling above this z-score in the upper tail, which is 0.9981.
