Question

MacBook Pro Problem 7: Is pl Two whether there is a difference o microprocessors are compared on a sample of eight benchmark codes to determine in speed. The times án seconds) used by each processor on each code are given in a the following table and saved FinalSP18.MTW Code 2 27.2 181 19.3 6 22.1 7 23.4 8 20.5 3 Processor A Processor B 4 19.7 5 24.5 27.6 20.1 I theft: 3.1 a) Can we conclude that the mean speeds of the two processors differ at o 0.05?

Answer 1

Differences (A - B) We calculate sample mean and std deviation from given data Sample Mean, d-Σ(d)--14.7 Sample Variance, s2 Sample std dev, s-Vs2v10.0826793.175323 -=-1.8375 = (d-d2 70.57875 A Test: Hd 0 with t dist" Hypothesis test (Null Hypothesis) (Alternative Hypothesis, also called Hı) This is two tailed test a =-1.8375 s 3.175323 Significance Level, a = 0.05 Degrees of freedom df-n-17 -1.8375-0 Test statistic,d3.175323/V8 Test statistic-1.637 Using t-table, we can observe that the p-value is between 0.10 and 0.20 Using excel we find p-value-Plt < I-1.637) = 2*t.dist(1.637, 7,1 ) = 0.1456937004 2.365 from t-table, two-tails, d.f.7) Critical Value, te = ± 2.365 Rejection criteria: Rejection region is in direction of Alternative hypothesis, since Ha έ 0, rejection rule is: Reject Ho if tte- 2.365 i.e, we reject if (t-2.365) or (t>2.365 equivalently if p-value < α Rejection region has been shaded

0.025 0.025 2.365 2.365 Decision: to S te or equivalently since p - value 2 a, we (fail to reject the null hypothesis. Hence at 5% significance level, there is insufficient evidence to conclude that μd is different from 0 The two processors do not differ significantly b) 95% CI for μd using t-dist Sample Mean-d1.8375 Sample Standard deviations 3.175323 Sample Size -n- 8 Significance level-α-1-0.95-0.05 Degrees of freedom for t-distribution, d.f.-n -17 Critical Value - ta/2,df to.025,df- 2.365 (from t-table, two-tails, d.f.-7) S d 3.175323 Margin of Error = E-ta/2,df × E= 2.365 × 1.122646 /8 Margin of Error, E 2.655058 Limits of 95% confidence interval are given by Lower limit-d-E =-1.8375-2.655058-e-4492558 ~-4.493 Upper limit -d+E-1.83752.6550580.8175580.818 95% confidence interval is: d士 - (-4.492558, 0.817558) 一1.8375士2.655058 95% CI using t-dist: -4.493 < Ha < 0.818 Note that exact answer using technology is: (-1.8375 3.17532338060686/SQRT (8) *T. INV (1-(1-0.95)/2,8-1 ,

-1.8375 3.17532338060686/SQRT (8) *T.INV (1-(1-0.95)/2,8-1) ) (-4.4921367791 , 0.8171367791) c) We assume that the differences are approximately normally distributed. And that samples are randomly drawn. We can test former assumption by drawing a QQ plot To draw QQ plot, we need to plot the ith ordered value against_th quantile of standard normal distribution n+1 i/(n+1) Obser vations Standard Normal Quantiles 0.222222 0.333333 0.444444 0.555556 0.666667 -3.40 -3.10 -2.40 1.20 0.40 0.40 3.10 1.221 -0.7648 -0.431 -0.1398 0.140 0.431 0.7648 1.221 0.888889 E -4 0.5 0.5 Observations Since sample quantile plot shows points reasonably close to a straight line, the normality condition is satisfied