with written work When morphine is injected, it typically causes decreased motor activity.However, we develop a tolerance to morphine with repeated dosages, so agreater dosage is needed to produce the same effect. We are testing a newdrug, Bloomingtine which may not have the same tolerance characteristics.We inject the same 5 rats with the same dosage of Bloomingtine every dayfor 4 days, and measure their motor activity. Go through the steps ofhypothesis testing to determine if motor activity changes over the 4different days (set a = .05). The data follow (higher numbers indicate greateractivity).RatDay 1Day 2Day 3Day 4P=A45719B335819C444618D236718E327820M=33.25.47.2G=94ΣΧT=15162736=514

Given information: Day 1 Day 2 Day 3 Day 4 A 3 4 5 7 B 3 3 5 8 C 4 4 4 6 D 2 3 6 7 E 3 2 7 8 Use the given data to form the excel table: Day 1 Day 2 Day 3 Day 4 Sum Mean A 3 4 5 7 19 4.75 B 3 3 5 8 19 4.75 C 4 4 4 6 18 4.5 D 2 3 6 7 18 4.5 E 3 2 7 8 20 5 Sum 15 16 27 46 94 Mean 3 3.2 5.4 7.2 4.7 D.FRows=a-1=5-1=4D.FColumn=b-1=4-1=3D.FError=a-1b-1=(5-1)·(4-1)=4×3=12SSA=b∑X¯i-X¯2=44.75-4.72+4.75-4.72+4.5-4.72+4.5-4.72+5.0-4.72=40.0025+0.0025+0.04+0.04+0.09=0.70SSB=a∑X¯j-X¯2=53-4.72+3.2-4.72+5.4-4.72+7.2-4.72=52.89+2.25+0.49+6.25=59.4SSAB=∑i=1a∑j=1bX¯ij-X¯i-X¯j+X¯2=3-4.75-3+4.72+4-4.75-3.2+4.72+...................8-5-7.2+4.72=12.1SST=SSA+SSB+SSAB=0.70+59.4+12.1=72.2MSA=SSADFA=0.704=0.175MSB=SSBDFB=59.43=19.8MSE=SSEDFE=12.112=1.0083FA=MSAMSE=0.1751.0083=0.17355372≈0.1736FB=MSBMSE=59.41.0083=19.6363636≈19.6364 Construct the Anova table using the obtained values: ANOVA Source of Variation SS df MS F P-value Rows 0.7 4 0.175 0.17355372 0.947745 Columns 59.4 3 19.8 19.6363636 0.0000637 Error 12.1 12 1.008333 Total 72.2 19 Null hypothesis: H0: μ1=μ2=μ3=μ4 Alternative hypothesis: HA : At-least one of the μ is not equal. For this analysis, the significance level is 0.05. Analyse sample data. F statistics is given by:- F = 19.64 Use the excel formula, "=F.DIST.RT(19.64,3,12)" to determine the P-value: The P-value = 0.00006 Interpret results. Since the P-value (0.00006) is less than the significance level (0.05), we have to reject the null hypothesis. Conclusion: Reject H0, There is sufficient evidence to conclude that the motor activity changes over 4 different days.