25.2) When plants are attacked by insects they release chemical compounds that attract other insects that prey on the at

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25.2) When plants are attacked by insects they release chemical compounds that attract other insects that prey on the attacking insects. The protective compounds that were released by the potato plant, Solanum tuberosumin, in response to attacks from various pests were examined in an experiment in Arizona. The investigators compared the amount of defensive compounds released by the potato plant while under the attack of four insects and a control plot that was unmolested by insect posts. The mean emissions of defensive compounds released by the potato plants in response to each treatment were measured over a period of 24 hours. The results are summarized in the table below. Use a one way ANOVA hypothesis at a 0.1% LOS with preliminary checks to determine if the mean rate of the compounds released in nanograms per hour (ng/hr) is significantly different for at least one of the attacking insects.

Treatment N Standard deviation Control Flea beetle CO potato beetle Wireworm Psyllid 21 23 27 11 16 compound emission rate (ng/hour) 14.1 28.3 24.8 21.1 23.2 1.9 2.7 2.8 2.2 2.9 25.2) Step 1a: Preliminary Checks • The data sets are collected from independent and random samples. . With summarized data, we assume that all data sets pass va normality test by the Anderson-Darling normality test, as we are using mean values.

25.2) Step 1b: Check for Homoscedastity (equal variances): K Because of unequal sample sizes we must first calculate a harmonic mean for the grouped sample size, n'. n = 1 TI + +. + n n' Use at least second decimal accuracy n Rounded up to the next Integer for f-table use. Establish the decision line for Hartley's Equal variance test. Recognize that this is NOT the decision line for the ANOVA test, rather it is the decision line for the preliminary check of equal variances, without which we can not proceed with the ANOVA test For Hartley's tost: df = (k, n-1)-( Do Not Reject Reject Fcritical

25.2) Step 1c: The statistical conclusion for Hartley's F test is: At the 5% LOS We Do Not Reject the null hypothesis of H., because: F-Notational Support: Fsample < Fcritical Numeric Validation: D-Notational Support: p> να Numeric (p > 0.20) 0.05 Validation: With the preliminary checks of independent and random samples, a normal distribution and equal variances, all met, we can proceed with a one-way ANOVA test.

25.2) Step 2: Declare the variables for this hypothesis test: ol English text equivalence for the Variable and Hypothesis Options: Multi-parameter Hypotheles: Note! Because of the limitations in A x = x3 = t; = X = is Variable English software most special characters B Ry # 2 2 ² 2 ₂ ² 2 ² ks Equivalence cannot be included in a pop-up At least one mean is not equal answer format and yet they are X; x-bar-i D needed for the hypothesis formation E Hi The pop-up selections will be in the My & H₂ H₂ H ² Hs mu-1 text-form for the special characters Byty = 1; -*t's embedded within the pop-up answer X X-tilde-i G R 2₂ # 2 2 2 Use the table at the right as a guide if HAt least one median is not equal mi eta i needed 11-16 1:11. ns 25.2) Step 2: Declare the variables for this hypothesis tout: muri - The mean emission rate in ing/hour) for a potato plant responding to an attack from the il insect F 1 a 25.2) Step 3) Choose the correct hypothesis statement to test the claim that the mean emission rate in ing noun for a potato plant varies depending upon the type of attacking insect. Tost the hypothesin at a 0.1% LOS H. H:C O= 0.001

25.2) Step 4) Complete the partially filled ANOVA table as an aid in forming the conclusion for the ANOVA hypothesis. ANOVA TABLE Source Sum of Squares (SS) Degrees of freedom: df dfs=1 Mean Squares (MS) F-sample SSB = 2425.45 MSB = FSample = Between Groups Within Groups SSW = df MSW = 6.57 Bracketed p-value: Total SST df- (p <0.0001)

25.2), Step 5) Write a statistical conclusion for the ANOVA hypothesis. At the 0.1% LOS we Reject the null hypothesis of H., because: df and F Notation Support: df = (dfs, dfw) = (0,0 F sample Fcritical Numeric Validation: > p-Notational Support: p < va Numeric Validation: p = (p <0.0001) V 0.001 25.2), Step 6) Write an English sentence conclusion The evidence supports the case that mean rate of emissions in (ng/hour) for the potato plant varies according to which insect is attacking the plant.