Asprinkler manufacturer claims that the average activating temperatures is at least 132 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.

Step-by-step explanation:

Given that a sprinkler manufacturer claims that the average activating temperatures is at least 132 degrees.

Sample size n =32:

Sample mean x bar =133

population std dev = sigma = 3.3 degrees.

Create null hypotheses:

(One tailed test)

Mean difference =

Since sigma is known we can use z test.  

Std error = sigma/sqrt of sample size =

Test statistic = Mean difference/Std error = 1.7142

p value = 0.0433

i think is 3 and 6. this took me 10 min to think.hope to and hope that my answer is brainliest.(follow me )

8 2/3 is the answer

step-by-step explanation: just use a calculator