Question

A sprinkler 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.

Answer 1

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:

`H_0: x bar = 132\\H_a: x bar \geq 132`

(One tailed test)

Mean difference =
`133-132=1`

Since sigma is known we can use z test.

Std error = sigma/sqrt of sample size =
`(3.3)/(√(32) ) \\=0.5834`

Test statistic = Mean difference/Std error = 1.7142

p value = 0.0433