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

Answer:

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:

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

(One tailed test)

Mean difference = $133-132=1$

Since sigma is known we can use z test.  

Std error = sigma/sqrt of sample size = $\frac{3.3}{\sqrt{32} } \\=0.5834$

Test statistic = Mean difference/Std error = 1.7142

p value = 0.0433