For two-way classification, let number of treatments and blocks be 5 and 5, respectively. It is given that SSTR (sum of squares due to treatments) = 18.5335, SSE (sum of squares due to error) = 3.545 and SST (total sum of squares) = 46.9335. The value of F-statistic for testing equality of effectiveness of five treatments is:
20.9
The F-statistic is calculated as:
F = (SSTR / (k - 1)) ÷ (SSE / (n - k))
Here, k = 5, n = total observations = 5 × 5 = 25
F = (18.5335 / 4) ÷ (3.545 / 20) = 4.633375 ÷ 0.17725 ≈ 20.9.
Important Questions from Hypothesis testing
If α is the level of significance and if (1 − α) is increased, then the width of the confidence interval of mean:
The analysis of variance technique was introduced by:
The power of a test is:
The term ‘Analysis of variance’ was introduced by:
Which of the following can be applied as a goodness-of-fit test?
