Question

Data:

5. You have collected data on the processing speed for 20 super computers, both before and after a newly designed chip was implemented. (a) Construct and interpret an 80% confidence interval for the difference between the average com- puting speed before and after the implementation of the new chip. Provide support for both your choice of interval and the validity of the underlying assumptions. (b) Provide the R code which verifies your findings in (a).

"","x"
"1",0.0458835170914729
"2",0.186357728205621
"3",0.261930461196287
"4",0.121678145570879
"5",0.190702709369361
"6",0.244160178413892
"7",1.12384053562444
"8",1.76960612885786
"9",0.360237084315125
"10",0.164134511258453
"11",0.209619600170602
"12",0.195104011489699
"13",0.163665047740577
"14",0.826605167818777
"15",0.523335928069361
"16",0.971550465234638
"17",0.664677629840183
"18",0.264826960157033
"19",0.383002542661351
"20",0.139428810965134

Answer 1

page-) Given that, collected data on the processing speed for 20 super compatess. so let us - nis sample size. - ad is mean of the differance. - Scd = standard deviation of the debterance, i na90 h = 0.440517. (a). Confidence interval for the difference between the average computing speed and after the omplementation of the new chip is = 80%. - Xd I 0:20 x 19 x 5 0.43373 -x19-1.7 - 0:440517 + 1.729137 120 = 0.440517 I 0167699. (0.272818, 0.608216)

page 2 from the above calcalations we conclude that, confidane onrual i for the before and after Omplementation for the a of the new chip is does not include the value is zero (o). Mence, he chip at 0.20 significance level, The computing speed is reduced afte, the emplemantation of re new chip. Histogram of x 0.0 0.44 0.8 1.2 1.6

fone 3 The histagram of the differance value is Positively skew, so the assumption of the normal distribution on the diffesance value & does nod satisfied. Therefore, the aove confidone Onterval may be invalid. (6) The R code which verties your findings in (a), Given mat Data x = c(0.45854, 0.186358, 0.26193, 0-121678, 0.190703,0:24416 1.123841, 1.769606, 00360237 On 164135, 0.20962, 0.195104, o 163665 6-826605, 0.52333, 0.97155 0-664678,6-264827 0.383003, 4394 29). ta wit us find mean value. - I mean = mean (x). s=sd(x) - no longth (x)=20.

page-4 - t= qt (P-o.10, df=19, low. de 19, lower fail of) [1] 1.327728 =) tq (p=orlo, df=19 power fail=f) [1] 1.327728 - 80% confidance onlerval / sast - lower-enterval= tnean-tox -) lower - Interval [] 0.3117484 - upper - Interval= Xmeant t xs/sart. - upper - Interval [ 0.5692804