Question

Three different washing solutions are being compared to study their effectiveness in retarding bacteria growth in five-gallon milk containers. The analysis is done in a laboratory, and only three trials can be run on any day. Because days could represent a potential source of variability, the experimenter decides to use a model that reduces experimental error (and hence increases precision) by grouping days into blocks. Observations are taken for four days, values in table below show measurement of effectiveness (bacteria growth rate after using the solution) for each day and solution type.

Solution 1 Days 3 18 17 1 2 22 24 4 13 16 5 4 39 44 22 2 3

Analyze the data from this experiment (use significance level α = 0.05) using an appropriate model and draw conclusions.

Answer 1

The output of two way anova factor is

A B c D E F K M solution G H Anova: Two-Factor Without Replication 1 2 3 4 1 13 22 18 39 16 24 44 2 3 17 1 Sum 92 5 4 22 4 SUMMARY Count 1 2 3 Average Variance 23 127.3333 25.25 168.9167 8 90 4 101 4 32 1 3 3 2 34 11.33333 32.33333 50 16.66667 121.3333 36 12 91 3 3 4 3 105 35 133 df ANOVA Source of Varia Rows Columns Error SS 703.5 1106.917 51.83333 MS F P-value Fcrit 2 351.75 40.71704 0.000323 5.143253 3 368.9722 42.71061 0.000192 4.757063 6 8.638889 Total 1862.25 11

Sola The null mean and alternative hypothesis is Ho: There is no difference between three Washing solutions. Ha: There is difference between three washing solutions. mean P-value = 0.0003 since P-value <<0.05), hence we reject the Ho. There is sufficient evidence to conclude that difference there is the washing between solutions. mean