STEM Problem Detail

Problem Statement

Prompt

Provide the solution of the Example 5.6 utilizing Table 5.26

Solution Rendering...

## Solution to the Birth-Weights Example

Using the 5 × 5 table of observations, we treat

* **Factor A** = *order of gravida* (5 levels),
* **Factor B** = *age-group of mother* (5 levels),

with **3 observations per cell**. The table matches the standard transcription of this example.

**Null hypotheses:**

\[
H_{0AB}:\ (\alpha\beta)_{ij}=0 \quad \text{for all } i,j
\]
\[
H_{0A}:\ \mu_{1\cdot}=\mu_{2\cdot}=\mu_{3\cdot}=\mu_{4\cdot}=\mu_{5\cdot}
\]
\[
H_{0B}:\ \mu_{\cdot1}=\mu_{\cdot2}=\mu_{\cdot3}=\mu_{\cdot4}=\mu_{\cdot5}
\]

So we test the **interaction first**, and then the two main effects.

### Step 1: Totals

\[
N=5\times 5\times 3=75
\]

Row totals ($$T_{i..}$$):

\[
T_{1..}=75.2,\quad T_{2..}=79.6,\quad T_{3..}=86.8,\quad T_{4..}=88.9,\quad T_{5..}=89.2
\]

Column totals ($$T_{.j.}$$):

\[
T_{.1.}=84.0,\quad T_{.2.}=86.7,\quad T_{.3.}=84.2,\quad T_{.4.}=82.0,\quad T_{.5.}=82.8
\]

Grand total:

\[
G=419.7
\]

Also,

\[
\sum\sum\sum x_{ijk}^2=2362.11
\]

### Step 2: Correction Factor

\[
\text{C.F.}=\frac{G^2}{N}=\frac{(419.7)^2}{75}=2348.6412
\]

### Step 3: Total Sum of Squares

\[
\text{T.S.S.}=\sum\sum\sum x_{ijk}^2-\text{C.F.}
\]

\[
\text{T.S.S.}=2362.11-2348.6412=13.4688
\]

### Step 4: Sum of Squares for Gravida (Factor A)

\[
\text{S.S.A.}=\left(\frac{\sum T_{i..}^2}{br}\right)-\text{C.F.}
\]

Here, $$b=5$$ and $$r=3$$, so $$br=15$$.

\[
\text{S.S.A.}=\frac{75.2^2+79.6^2+86.8^2+88.9^2+89.2^2}{15}-2348.6412
\]

\[
\text{S.S.A.}=10.3781
\]

### Step 5: Sum of Squares for Age-group (Factor B)

\[
\text{S.S.B.}=\left(\frac{\sum T_{.j.}^2}{ar}\right)-\text{C.F.}
\]

Here, $$a=5$$ and $$r=3$$, so $$ar=15$$.

\[
\text{S.S.B.}=\frac{84.0^2+86.7^2+84.2^2+82.0^2+82.8^2}{15}-2348.6412
\]

\[
\text{S.S.B.}=0.8501
\]

### Step 6: Sum of Squares for Cells

Let $$T_{ij.}$$ be the total for the $$(i,j)$$-th cell. Then

\[
\text{S.S.}_{\text{cells}}=\left(\frac{\sum\sum T_{ij.}^2}{r}\right)-\text{C.F.}
\]

\[
\text{S.S.}_{\text{cells}}=11.7688
\]

So the interaction sum of squares is

\[
\text{S.S.AB}=\text{S.S.}_{\text{cells}}-\text{S.S.A.}-\text{S.S.B.}
\]

\[
\text{S.S.AB}=11.7688-10.3781-0.8501=0.5405
\]

### Step 7: Error Sum of Squares

\[
\text{S.S.E.}=\text{T.S.S.}-\text{S.S.A.}-\text{S.S.B.}-\text{S.S.AB}
\]

\[
\text{S.S.E.}=13.4688-10.3781-0.8501-0.5405=1.7000
\]

### Step 8: ANOVA Table

\[
\begin{array}{|l|c|c|c|c|}
\hline
\textbf{Source of Variation} & \textbf{d.f.} & \textbf{S.S.} & \textbf{M.S.} & \textbf{F} \\
\hline
\text{Gravida (A)} & 5-1=4 & 10.3781 & 2.5945 & 76.31 \\
\hline
\text{Age-group (B)} & 5-1=4 & 0.8501 & 0.2125 & 6.25 \\
\hline
\text{Interaction (A×B)} & (5-1)(5-1)=16 & 0.5405 & 0.0338 & 0.99 \\
\hline
\text{Error} & 75-25=50 & 1.7000 & 0.0340 & \\
\hline
\text{Total} & 75-1=74 & 13.4688 & & \\
\hline
\end{array}
\]

### Step 9: Decision

At the 5% level of significance,

\[
F_{0.05}(4,50)\approx 2.56,\qquad F_{0.05}(16,50)\approx 1.85
\]

Since

\[
F_{AB}=0.99<1.85,
\]

the **interaction is not significant**.

Also,

\[
F_A=76.31>2.56,\qquad F_B=6.25>2.56,
\]

so both **order of gravida** and **age-group of mother** have a **significant effect** on birth-weight.

### Conclusion

\[
\boxed{\text{The birth-weight is significantly affected by both the age of the mother and the order of gravida.}}
\]

\[
\boxed{\text{The interaction between the two factors is not significant.}}
\]

Loading solution...

Problem #83