Design of Experiment

Design of Experiments                  Homework # 01

2.6 Suppose that we are testing H0: µ=µ0 vs. H1: µ>µ0 with a sample size of n=15. Calculate bounds on the P-value for the following observed values of the test statistics: (10 POINTS)

(a) t0=2.35

(b) t0=3.55

(c) t0=2.00

(d) t0=1.55

(a)        t0 = 2.35        Table P-value = 0.01, 0.025        Computer P-value = 0.01698

(b)        t0 = 3.55         Table P-value = 0.001, 0.0025        Computer P-value = 0.00160

(c)        t0 = 2.00         Table P-value = 0.025, 0.05        Computer P-value = 0.03264

(d)        t0 = 1.55         Table P-value = 0.05, 0.10        Computer P-value = 0.07172

2.8 Consider the Minitab output shown below: (10 POINTS)

(a)        Fill in the missing values in the output.  Can the null hypothesis be rejected at the 0.05 level?  Why?

Std. Dev. = 2.3365        UCI = 93.5450

Yes, the null hypothesis can be rejected at the 0.05 level because the P-value is much lower at 0.002.

(b)        Is this a one-sided or two-sided test?

Two-sided.

(c) If the hypotheses had been : µ=90 vs. H1: µ≠90 would you reject the null hypothesis at the 0.05 level?

Yes.

(d)        Use the output and the t table to find a 99 percent two-sided CI on the mean.

CI = 91.2735, 93.8875

(e)        What is the P-value if the alternative hypothesis is H1: µ > 91?

P-value = 0.001.

2.20 Two types of plastic are suitable for use by an electronic calculator manufacturer.  The breaking strength of this plastic is important.  It is known that 1 = 2 = 1.0 psi.  From random samples of n1 = 10 and n2 = 12 we obtain [pic 1]1 = 162.5 and [pic 2]2 = 155.0.  The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi.  Based on the sample information, should they use plastic 1?  In answering this questions, set up and test appropriate hypotheses using  = 0.01.  Construct a 99 percent confidence interval on the true mean difference in breaking strength. (10 POINTS)