Question 6. Does home access to Web-based monitoring and graphical feedback conc

Question

Question 6. Does home access to Web-based monitoring and graphical feedback concerning nutrition during pregnancy improve nutrition in comparison with no access, or Web-based monitoring without graphical feedback? An experiment was conducted and the following data from three random samples of pregnant female subjects were collected. Data were measured on a composite standardized nutritional index. Conduct a one-way analysis of variance to answer the question. Please follow the 5-step hypothesis testing procedure and fill in the table below (shaded in red). (In real research investigations, such sample sizes are small. Here, please focus on understanding the concept and procedure.) No Access: 2, 8, 4, 6 Access: no graphics: 1, 4, 7, 5, 6 Access: graphics: 2, 9, 6, 1, 1 Find Sums of squares (SS), df, Mean Squares (MS), F, and Sig. for: Source of Variation: A.) Between, B.) Error(or Residual), and C.) Total

Explanation / Answer

Step 1

Null Hypothesis Ho :µ1 =µ2 =µ3

Alternative Hypothesis : one of the means is diffrent among

Step 2

Degrees of freedom between = k - 1 = 3 - 1 = 2

Degrees of freedom Within = n - k = 14 - 3 = 11

Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.982

Step 3

Grand Mean = G / N = 4.4

SST = ( Xi - GrandMean)^2 = 95.43

SS Within = (Xi - Mean of Xi ) ^2 = 92.00

SS Between = SST - SS Within = 3.43

Step 4

Mean Square Between = SS Between / df Between = 1.714

Mean Square Within = SS Within / df Within = 8.364

Step 5

F Cal = MS Between / Ms Within = 0.20

We got |F cal| = 0.20 & |F Crit| =3.982

MAKE DECISION

Hence Value of |F cal| < |F Crit|and Here We Accept Ho

Mean n Std. Dev 5.0 4 2.58 No Access 4.6 5 2.30 Access: no graphics 3.8 5 3.56 Access: graphics 4.4 14 2.71 Total

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