Reject H0. At the significance level of 5%, there is enough evidence to conclude that there is a significant difference in the mean breaking strength of the 4 cables.
As pergiven data:
No. of treatments (k) is 4
Total sample size (nt) is 24
Sum of Squares, total (SST) is 85.05
Sum of Squares, treatment (SSTR) is 61.64
Significance level (\alpha) is 0.05
Hypothesis testing is carried out as follows,
Null hypothesis, H0 : \mu1 = \mu2 = \mu3 = \mu4
Alternative hypothesis, H1 : Not all the population means are equal
Sum of Squares, error (SSE)
= SST-SSTR
= 85.05-61.64
= 23.41
Mean Squares, treatment (MSTR)
= SSTR/(k-1)
= 61.64/(4-1)
= 20.5467
Mean Squares, error (MSE)
= SSE/(nt-k)
= 23.41/(24-4)
= 1.1705
Test statistic, F
= MSTR/MSE
= 20.5467/1.1705
= 17.554
Numerator degrees of freedom (df1)
= k-1
= 4-1
= 3
Denominator degrees of freedom (df2)
= nt-k
= 24-4
= 20
Source SS df MS F
Treatment 61.64 3 20.5467 17.554
Error 23.41 20 1.1705
Total 85.05 23
From the F-table,
At df1 = 3, df2 = 20, \alpha = 0.05 and right-tailed,
Critical value, Fc = 3.098
Since F (i.e.,17.554) > Fc (i.e.,3.098), so we reject H0.