Question

5. 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 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4] b) Find a 95% CI for then true average system-activation temperature. [3]

Answer 1

Step 1:

Ho : $\mu$ = 130 (Claim)

Ha: $\mu$ $\neq$ 130

Step 2: test statistics

n = 9

sample mean = 131.08

population sd = 1.5

Assuming that data is normally distributed and as population sd is given, we will use z stat

$z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{131.08 - 130 }{1.5/\sqrt{9}}$

z = 2.160

Step 3: Rejection region:

alpha = 0.01

The z-critical values for a two-tailed test, for a significance level of α=0.01

zc = − 2.58 and zc = 2.58

Reject Ho if z < -2.58 or z > 2.58

Step 4: Conclusion:

As the z stat does not fall in the rejection region, we fail to reject the Null Hypothesis.

Step 5: Decision

Hence we do not have sufficient evidence to believe that the activation time is not equal to 130

Data does not contradict the claim at 0.01 level.

(B) Confidence interval

CI = mean +/ E

$E = z * \frac{\sigma}{\sqrt{n}} = 1.960 * \frac{1.5}{\sqrt{9}} = 0.98$

CI = 131.08 +/- 0.98

CI = 130.10 , 132.06

As 130 does not fall in the CI range, we reject the Null hypothesis at 95% confidence level.