Solution 3 of 4 You were asked to make a decision, given the following information: Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is \( 4.8 \) parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 16 samples is \( 4.4 \) ppm with a variance of \( 0.64 \). Assume the population is normally distributed. A level of significance of \( 0.05 \) will be used. Make the decision to reject or fail to reject the null hypothesis. The \( P \)-value is the probability of observing a value of the test statistic as extreme or more extreme than the one observed in the data, assuming that the null hypothesis is true. If we are using technology, we want to find \( P(|t| \geq 2) \) for the \( t \)-distribution with, \( d f=16-1=15 \). So the exact \( P \)-value, rounded to four decimal places, is \( 0.0639 \). If we are using the table of \( t \)-critical values, then we want to find the critical values for the area in two tails for the \( t \)-distribution with 15 degrees of freedom. Since the \( t \)-distribution is symmetric, we want to compare the critical values with the absolute value of the test statistic: 2 . The two critical values that lie on either side of the test statistic create an interval for the \( P \)-value from the smaller area to the larger area, which is \( (0.05,0.1) \).