Question

three different traffic routes are tested for mean driving time. the entries in the table 13.18 are the driving times in minutes on the three different routes.
route 1 route 2 route 3
30 27 16
32 29 41
27 28 22
35 36 31
table 13.18
state f statistic.
a. 0.2653
b. 0.8791
c. 1.5792
d. 8.8862

Explanation:

Step1: Calculate the group means

For Route 1: \(\bar{x}_{1}=\frac{30 + 32+27+35}{4}=\frac{124}{4} = 31\)
For Route 2: \(\bar{x}_{2}=\frac{27+29+28+36}{4}=\frac{120}{4}=30\)
For Route 3: \(\bar{x}_{3}=\frac{16 + 41+22+31}{4}=\frac{110}{4}=27.5\)

Step2: Calculate the grand mean

\(\bar{x}=\frac{124 + 120+110}{12}=\frac{354}{12}=29.5\)

Step3: Calculate the sum of squares between groups (SSB)

\(SSB=4\times(31 - 29.5)^{2}+4\times(30 - 29.5)^{2}+4\times(27.5 - 29.5)^{2}\)
\(=4\times(2.25)+4\times(0.25)+4\times(4)\)
\(=9+1 + 16=26\)

Step4: Calculate the sum of squares within groups (SSW)

For Route 1: \((30 - 31)^{2}+(32 - 31)^{2}+(27 - 31)^{2}+(35 - 31)^{2}=1 + 1+16 + 16=34\)
For Route 2: \((27 - 30)^{2}+(29 - 30)^{2}+(28 - 30)^{2}+(36 - 30)^{2}=9+1+4 + 36=50\)
For Route 3: \((16 - 27.5)^{2}+(41 - 27.5)^{2}+(22 - 27.5)^{2}+(31 - 27.5)^{2}=132.25+182.25+30.25 + 12.25=357\)
\(SSW=34+50 + 357=441\)

Step5: Calculate the degrees of freedom

Degrees of freedom between groups \(df_{B}=k - 1=3 - 1=2\) (where \(k = 3\) is the number of groups)
Degrees of freedom within groups \(df_{W}=n - k=12 - 3=9\) (where \(n = 12\) is the total number of observations)

Step6: Calculate the mean square between groups (MSB) and mean square within groups (MSW)

\(MSB=\frac{SSB}{df_{B}}=\frac{26}{2}=13\)
\(MSW=\frac{SSW}{df_{W}}=\frac{441}{9}=49\)

Step7: Calculate the F - statistic

\(F=\frac{MSB}{MSW}=\frac{13}{49}\approx0.2653\)

Answer:

A. 0.2653