help please

5. Eighteen plants of the same age and genetic strain were randomly divided into three groups, each containing 6 plants. One group was left untreated, one was infected with virus A, and one was infected with virus B. After one week, the activity of a particular enzyme (in pl 02/ mg protein/ min) was measured in cach plant. The experiment was conducted to investigate whether infection by either virus affects the activity of this enzyme. The following data were obtained: Control Virus A Virus B 1.47 2.44 2.87 1.62 2.31 3.05 1.06 1.98 2.36 0.89 2.76 3.21 1.67 2.39 3.00 1.21 2.54 3.10 State the response variable, the factor and the treatments under investigation, experimental units and number of replicates per treatment. [4] Use the exploratory data analysis provided in the MINITAB output to indicate if there are trends in the data to support the research hypothesis. [4] At a 5% level of significance test for a difference in mean enzyme activity among treatments. Use the appropriate MINITAB output to report the results of this test. Provide full details including hypotheses, test statistic and p-value, test decision and conclusion. [7] (iv) State the assumptions of the hypothesis test used under (ili) and investigate whether these assumptions are satisfied. Use the MINITAB output as appropriate, give the name of the tests, provide the test statistic and p-values, interpret any graphical output and formulate the conclusion. [5] (v) Use MINITAB output to investigate which treatment has produced a difference, include the name of the test, the measures reported and any relevant results with interpretation. [5]05 MINITAB output: Statistics Variable Treatment Mean SE Mean StDev Q1 Median Q3 Enzyme_activity Control 1.320 0.129 0.316 1.018 1.340 1.633 VirusA 2.403 0.106 0.259 2.228 2.415 2.595 VirusB 2.932 0.123 0.302 2.742 3.025 3.128 Boxplot of Enzyme_activity Treatment Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Treatment 2 8.100 4.05022 47.09 0.000 Error 15 1.290 0.08600 Total 17 9.390 Tukey Simultaneous Tests for Differences of Means Difference of Difference SE of Simultaneous Adjusted Treatment Levels of Means Difference 95%% CI T-Value P-Value VirusA - Control 1,083 0.169 (0.644. 1.523) 6.40 0,000 VirusB - Control 1.612 0.169 (1.172, 2.051) 9.52 0.000 VirusB - VirusA 0.528 0.169 (0 069, 0.968) 3.12 0.018 Individual confidence level = 97.97%6Residual Plots for Enzyme_activity Normal Probability Plot Versus Fit TO Find Value HiMogiram Vernon Order H -04 Paidund Probability Plat of 8651 UFI Test for Equal Varianou: Frayme_activity v Treatment