Question

A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system activation temperature is 130 ◦F. A sample of n = 49 systems, when tested, yields a sample average activation temperature of 131.08 ◦F. If the distribution of activation times is Normal with standard deviation of 1.5◦F, does the data contradict the manufacturer’s claim at significance level α = 0.01?

Answer 1

Solution :

The null and alternative hypothesis is ,

= 130  

= 131.08

= 1.5

n = 49

This will be a two tailed test because the alternative hypothesis is

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :   = 130

Ha :    130

Test statistic = z

= ( - ) / / n

= (131.08 - 130 ) / 1.5 / 49

= 5.04

P - value = 2* p (Z >5.04 ) = 2* 1 - P (Z < 5.04 )

= 2 * 1 - 1.0000

= 2 * 0.0000

P-value = 0.0000

= 0.05  

0.0000 < 0.05

P-value <

Reject the null hypothesis .

There is sufficient evidence to claim