Asprinkler manufacturer claims that the average activating temperatures is at least 135 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.

z-value = -3.4283

p-value = 0.0003

Step-by-step explanation:

To find the standardized test statistic or z-value, we use the formula

where

N = size of the sample.

So,

As your sampling suggests that the real mean could be less than the manufacturer's mean, then you are interested in the area under the normal curve to the left of -3.4283 and this would be your p-value.

We compute the area of the normal curve for values to the left of -3.4283 either with a table or with a computer and find that this area is equal to 0.0003.

So the p-value is  

step-by-step explanation:

d no solution

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