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

a)

$\\\mu_0 = 10 \\n = 9 \\s = 1.5 \\\bar{X} = 131.08 \\\alpha = 0.01$

The test hypothesis is

$\\\text{Null Hypothesis }--> H_0: \mu = \mu_0 \\\text{Alternate Hypothesis }--> H_1: \mu \ne \mu_0$

This is a two-sided test because the alternative hypothesis is formulated to detect hypothesized mean value on either side.

Now, the value of test static can be found out by following formula:

X - PO to s/n 131.08 - 10.0 to 1.5/9 to = 242.16 Critical value = ta/2,n–1 = to.0025,8 = 3.3554 Rejection Region: to > ta/2,1-1 or to <-ta/2,n-1 For a 0.01, ta/2,n-1 t0.005,8 3.3554. Since to 242.16 > 3.3554 -0.005, we reject the null hypothesis Ho : j = 10.0 in favor of the alternative hypothesis H1 : u + 10.0 at a= = 0.01 ..

We have insufficient evidence to claim that the true average system activation temperature is 130 F

b)

Mean ) 131.08 Sample size (n) = 9 Standard deviation (s) 1.5 Confidence interval (in %) = 95 ta/2,n–1 = t0.025,8 2.306 Since we know that S Confidence interval = ī ta/2,n-1 Vn Required confidence interval (131.08 – 2.306 1.5 1.5 131.08 + 2.306-

Required confidence interval = (131.08-2.306(0.5), 131.08+2.306(0.5))

Required confidence interval = (131.08 - 1.153, 131.08 + 1.153)

Required confidence interval = (129.927, 132.233)

Interpretion: We are 95.0% confident that the true mean of the population lie between the interval 129.927 and 132.233.

Please hit thumbs up if the answer helped you.