Question

A sprinkler 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.

Answer 1

z-value = -3.4283

p-value = 0.0003

Explanation:

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

`z=(\bar x-\mu)/(\sigma/\sqrt N)`

where

`\bar x=mean\;you\;got`

`\mu=mean\;of\;the\;manufacturer`

N = size of the sample.

So,

`z=\frac{133-135}{3.3/\sqrt {32}}=-3.4283}`

`\boxed {z=-3.4283}`

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

`\boxed {p=0.0003}`