A sprinkler manufacturer claims that the average activating temperatures is at most 131 degrees. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133 degrees. Assume the population standard deviation is 3.3 degrees. Find the standardized test statistic and the corresponding p-value.

Group of answer choices

z-test statistic = 3.43, p-value = 0.0006

z-test statistic = -3.43, p-value = 0.0006

z-test statistic = -3.43, p-value = 0.0003

z-test statistic = 3.43, p-value = 0.0003