Math 3070 - 1           Fourth Quiz SOLUTION               Treibergs 

Friday, November 30, 2001

A manufacturer of sprinkler systems used for fire protection in
university buildings claims that the true average system activation
temperature is 130 F.  A sample of n=16 systems, when tested yielded a
sample average activation temperature of 130.33 F.  If the
distribution of activation times is normal with a population standard
deviation of 1.50 F,  does the data contradict the manufacturer's
claim at significance level alpha=.05?

a.  State the null and alternative hypotheses. State a formula for the
test statistic. State the acceptance and rejection regions for the
null hypothesis. Compute the statistic and draw a conclusion.


H0: \mu  =  130  =  \mu_0

H1: \mu \ne 130  (not equal: TWO-TAILED TEST)

\sigma=1.50

Population standard deviation is given. Thus we may use the  Z-test.


z = ( \bar x - \mu_0 ) / ( \sigma / \sqrt{n} }



The critical score for a two-tailed test is z_a = z_{.025} = 1.960.
Thus we accept H0 if |z| <  z_{.025} = 1.960;
    and reject H0 if |z| >= z_{.025} = 1.960.

n=16

\bar x = 130.33

Hence

z = ( \bar x - \mu_0 ) / ( \sigma / \sqrt{n} }
  = ( 130.33- 130.00 ) / ( 1.50 / \sqrt{ 16} ) = .880.

Conclusion: Accept H0:  Not enough evidence at the 95% level to reject
                        the claim that \mu = 130.





b.  Determine the P-value. What conclusions do you draw from the
P-value at this level of significance?


For two-tailed tests, the P-value is

P = 2 P( Z < -|z|)
  = 2 P( Z < -.88) = 2 (.1894) = .3788.

The evidence that the mean activation temperature isn't 130 F is not
conclusive at all. The null hypothesis would be rejected only at
\alpha=.38 level. Not close to \alpha=.05.







c.  Find a 95% confidence interval for the mean activation
temperature. How does your confidence interval relate to your test of
hypothesis from part a?

The \alpha=.05 confidence bound on the mean is

\bar x - z_a \sigma/ \sqrt{n} < \mu < \bar x - z_a \sigma/ \sqrt{n}

Numerically,

     130.33 - (1.96)(1.50)/4 < \mu < 130.33 + (1.96)(1.50)/4 ,

                       129.60 < \mu < 131.07.


The null hypothesis is accepted if and only if \mu_0 falls within the
\alpha confidence interval for the mean. In this case \mu_0=130.00,
which is within the confidence interval, thus the null hypothesis is
accepted.