4 Suppose you are interested in studying whether scent affects spatial reasoning abilities. You decide to test spatial reasoning using completion timescores for the paper-folding test with five people, repeating the test on each person with three different scents (cinnamon, lavender, and peppermint).In this experiment, the null hypothesis is that:O There are no differences in the mean completion times among the scents comparedO There are no individual differences in the completion time meansO The completion time mean for at least one scent is different from anotherThe factor of interest is the, and the dependent variable is theThe results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, recorded inseconds.ScentParticipantCinnamonLavenderPeppermintParticipant Totals3.413.403.49P = 10.303.293.473.25P = 10.01%3D3.703.713.63P = 11.04153.684.083.88P = 11.64G54.703.804.173.74P = 11.71ΣΧ2200.4864T = 17.88T= 18.83T = 17.99SS = 0.1858SS = 0.4866SS = 0.2335The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population meansare equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for themissing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on theorder in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between youranswers and the answer choices.)ANOVA TableSSdfMSFSourceBetween treatments0.0755Within treatments0.1973Between subjectsError14Total SS = 0.1858The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population meansare equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for themissing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on theorder in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between youranswers and the answer choices.)ANOVA TabledfMSFSourceBetween treatments0.0755Within treatmentsBetween subjects0.1973ErrorTotal14F DistributionNumerator Degrees of Freedom -26Denominator Degrees of Freedom -265000500012481.00FUse the Distributions tool to find the critical value for a = 0.01. The critical value is F =At a significance level of a = 0.01, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis isYouconclude that scent affects spatial reasoning abilities.

Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts. In case you require the unanswered parts also, kindly re-post that parts separately. From the given information, studying whether scent affects spatial reasoning abilities. Null Hypothesis: H0: There is no difference in the mean completion time among the three different scents. Correct option: There is no differences in the mean completion times among the scents compared. In the given situation, the completion times are compared among the factors scents. Hence, the factor is scents and the dependent variable is completion time. Thus, the factor of interest is the scents and the dependent variable is the completion time. There are 3 treatments. df for treatments is number of treatments-1=3-1=2 There are 5 participants. df for Between groups is number of participants-1=5-1=4 Error degrees of freedom: Error df=df for treatments*df for groups=2*4=8 Total degrees of freedom=14 Therefore, within treatments df=4*3=12 Consider, SSwithin treatments=MSwithin treatments*dfwithin treatments =0.0755*12 =0.906 SS Between subjects=MSBetween subjects*dfbetween subjects =0.1973*4 =0.7892 Total sum of squares: SSTotal=∑x2-G2N =200.4864-54.70215 =1.0137 Sum of squares between treatments: SSBetween Treatments=∑Ti2n-G2N =17.882+18.832+17.9925-54.70215 =0.108 Sum of squares for Error=Total sum of squares-Between treatments sum of squares =1.0137-0.108 =0.9057 Therefore, the anova table is, Source SS df MS F Between treatments 0.108 2 0.108/2=0.054 F=MSBetweenMSError =0.477 Within treatments 0.906 12 0.0755 Between subjects 0.7892 4 0.1973 Error 0.9057 8 0.9057/8=0.113213 Total 1.0137 12