Essay About P-Value And T0

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 = 0.071722.8 Consider the Minitab output shown below: (10 POINTS)One-Sample T:YTest of mu=91 vs. not=91VariableNMeanSE MeanStd. Dev.95% CITPY2592.58050.4673?(91.6160,?)3.380.002(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 = 93.5450Yes, 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)