Your SlideShare is downloading.
×

- 1. H.W#1 Name: Rabil Iglance Sikder ID: & This study is an experiment because the factor of interest (diets) is being manipulated by the researcher and the response variable (milk production) is being measured in order to assess the effect of the diets on milk production. This manipulation of the factor distinguishes this study from an observational study, where the variables are simply observed and recorded without manipulation. ** The response variable for this study is milk production, which is being measured daily over a 30-day period for each cow. & The factor under study is the diet, with five levels:A, B, C, D and E # The observational units in each design are the cows.
- 2. * @ The experimental units in each design are go Design 1: The farm, as each farm receives a unique diet and thedaily milk production of all a cows on the farm is being measured. Design 2: The individual cow , as each cow receives a unique diet and its daily milk production is being measured. & o Blocking is used for Design 1, as the farms are being used as blocks to control for sources of variability within groups (farms) ⑰& Replication is accomplished for both designs, although the level of replication is different for each design: * Design 1: There is a low level of replication, as each diet is only represented once across the a farms * Design 2: There is a higher level of replication, as each diet is represented 8 times across the cows within each farm. This allows for a more accurate estimate of the diet's effect on milk production-
- 3. & This study is an experiment. It is an experiment because the investigator randomly assigned the 10 bottles to Machines and another 10 bottles to Machine 2. This manipulation of the :edependent variable (Maching makes if an experiment. 28 Side by side plot for each machine is constructed and the file is attacked. & We can calculate the standard reviations the informula: ch - 1 For machine 1, the standard deviation is 91 = 0.0302769 ounce For machine 2, the standard deviation is & 2 = 0.0254981 ounce
- 4. The poor mandard dation is the Nf + (3x0.025) =0.0280 ounce * T o determine if there is significant my difference in fill volume, we can Near perform means on hypothesis testto compare the entire Data: Machine A Machine I 5 = 16.015 52=16.005 & = 0.0302765 52 = 0.0254981 n = 10 u2 = 10
- 5. -test: We know, to -5th-o Here, Si = G+(v - 1), -oof + (3x0.02,ha i sp = 0.0280 O = to oooo = 0.8788
- 6. Given, x = 0.0 C degrees offreedom: n, the - 2 = 18 Now, we would reject Itoim, = M2 if the numerical value of the test statistic to > to.025,18 = 2.101 or if to <-to.025, 18=-2.101. We find that to = 0.8783<t= 2.101 0.025,18 So, we cannot reject Ho irl, = M2 i =0.2G · :P-value will be = 0.28x2 - 0.8 :We compare p-value with <=0.03 We get, p-value? X.
- 7. So, we fail to reject the will hypothesis. So, Ho: r, = ut is supported. So, we can conclude that the mean difference in fill volume between the two types of machine are equal. * We use the formula, 5.-Ja-tak,mitmn2*Na Iee,-en =j.-.+tax,n+in => -0.072h1M, - M, = 0.0922 So, 35 percent confidence interval estimate on the difference in means extends from -0.0722 to 0.0922 ounce
- 8. Since, el.-1 = 0 is included in this interval, so the data support the hypothesis that u, =M2 ③ we can calculate the standard reviations the reformula: For 95 the standard deviation is, 31=2.0995643 KA For 100° the standard deviation is, 9 h = 1.6404266 KA
- 9. We will do the hypothesis test to see whether the higher backing temperature results in water with a lower mean photoresist thickness. Data: 95°C 100°C 5 = 3.367 52 = 6.847 5, = 2.099 S = 1.640 2 n = 8 nz = 8 -test: We know, to -5th-o Here, Si = G(x - 1s n, + 42 - 2 : + (x1.s0 8 + 8 - 2
- 10. v = 3.5477 =) Sp . 9p = 1.8832KA O = to iGO T - 2.6789 Given, x = 0.0 C degrees offreedom = 8+8-2 = 19 Now, we would reject Itoim, = M2 if the numerical value of the test statistic to > to.025,16 = 2.145 or if to(- to.025,1n = - 2.145 * We find that to = 2.67893 t = 2.145 0.025, 14 So, we would reject Ho :r, = M2 so, the mean is different I H. ir , Elle
- 11. D I 0.00S · O LO * E - 2.67532.6793 The P-value will be between 0.01 and 240,005= 0.01 ↓ P-value will be = 0.002X2 - 0 . 01 :We compare p-value with <=0.03 We get, p-value (x & So, we reject the well hypothesis, e. = Mr
- 12. So, we can say there is enough support to claim that the higher baking temperature results in waters with a lower mean photoresist thicknell. ③ we use the formula, 5.-Ja-tak,mitmn2*Na Iee,-en =j.-.+tax,n+in => - 2.8825x15(u,- M-= 5.0400 So, 35 percent confidence interval estimate on the difference in means to extends from 12.885x105 5,0400KA
- 13. We are 35% confident that the mean difference is between - 2.8825x10" and 2,0400 KA