For the below questions, refer to the following study and the SAS output below:Dexterity, as a measure of hand function, is an important component of a thorough hand evaluation. This is especially true in children, for whom the relationship between the commonly measured parameters of range of motion, sensation, and strength may not reflect actual functional ability. Gogola et al. (2013) conducted a study to document normative values from the Functional Dexterity Test (FTD) for typically developing children and to optimize test administration and interpretation. The FDT is a timed pegboard test consisting of 16 thick cylindrical pegs arranged in 4 rows of 4 pegs each. Patients turn over all pegs in a specified order by manipulating each peg in their hand. A total of 174 typically developing children aged 3 to 17 years participated in the study. Children completed the 16-peg FDT with either their dominant (n=105) or nondominant (n=69) hand, and elapsed time was recorded in seconds. Data were analyzed as 16/time, interpreted as FDT speed (pegs per second). Using a 0.05 significance level and the given computer output, you need to test the claim that the mean FDT speeds for dominant (1) and nondominant (0) hands differ significantly after adjusting for age (in years) by answering the questions that follow.

These data were analyzed using two different models. Results from the analysis are provided below labeled SAS Output 1 and SAS Output 2. To answer some of the questions that follow, you need to fill in some critical pieces of information that have been deleted.

\fSAS Output 1 Sum of Source OF Squares Mean Square F Value PC> F Model A 1.90029775 C F Age 13.74614560 3.74614560 302.77 Itl Intercept 0.2309333571 8 0.02726990 8.47 <.0001 age dominant b least squares means adjustment for multiple comparisons: tukey-kramer ho:lsmean lsmean speed ec> It 0.69828145 <.0001 output sum of source df squares mean square fvalue pr> F Model A B 1. 26693001 C <.0001 error d e corrected total r-square coeff var root mse speed mean source df type lll ss square f value bc> F Age 1 3.22634279 3.22634279 259.25 <.0001 dominant age standard parameter estimate error : value bc> It] Intercept 0.2264932870 8 0.04482257 5.05 <.0001 age dominant b>