Question

Question 8 2 pts A sprinkler manufacturer claims that the average activating temperatures is at least 134 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 O z-test statistic - -1.71, p-value-0.0432 O z-test statistic -1.71, p-value -0.0865 O z-test statistic -1.71, p-value- 0.0865 O z-test statistic - 1.71, p-value-0.0432

Answer 1

H0: Null Hypothesis: $\mu \geq$ 134

HA:Alternative Hypothesis: $\mu <$ 134

SE = $\sigma$/$\sqrt{n}$

= 3.3/$\sqrt{32}$ =0.5834

Test statistic is:
Z = (133 - 134)/0.5834 = - 1.7141

Table of Area Under Standard Normal Curve gives area = 0.4568

So,

P - value = 0.5 - 0.4568 =0.0432

So

Correct option:

z - statistic = - 1.71,   p - value = 0.0432