STA 6166 UNIT 3 Section 3 Exercises
 Welcome < Begin < Section 3 < Unit 3 Section 3 Answers
 To Ag and Env. Answers To Tox and Health Answers To Social and Education Answers To Engineering Answers

# Ag and Environmental

1. An experiment is run to examine two different methods methods of preserving onions. The researcher proposes to compare Method A to Method B, by storing a large number of onions using each method and after 6 months to classifying each onion as 'good' or 'defective'. The result of the experiment is as below:

 Experiment Result Method A Method B Good 134 207 Defective 16 33 Total number 150 240

a. Use these data to test whether the proportion of GOOD onions in each method are different from each other. Use a=.05.

H0: pA - pB = 0

HA: pA - pB not equal to 0

T.S.:

RR: Reject if |z| > za/2=z0.025 = 1.96

Conclusion: Do NOT reject the null hypothesis and conclude that the proportion of good onions is the same for the two methods. Note that from this we would also conclude that the proportion of bad onions should also be considered the same.

b. Calculate a 95% confidence interval for the difference tested above.

The 95% confidence interval is computed using the equation:

2. A second aspect of the study was to examine the effects of different fertilizer treatments on the incidence of blackleg (Bacterium phytotherum) on potato seedlings. Seedlings were randomly assigned to a fertilizer treatment and after three weeks classified as to whether it was contaminated by blackleg or free. The results for four treatment were as follows:

 Observed frequencies Blackleg Total 1.No fertilizer 16 101 2.Nitrogen only 10 95 3.Dung only 4 113 4.Nitrogen and dung 14 141

Use these data to determine if the presence of Blackleg is independent of the Fertilization method. Use a=.05.

These data represent a 2 by 4 contingency table but is presented in odd format. More commonly you would expect to see a table like the one below.

 Disease Status Blackleg Clean Fertilization No Fertilizer 16 85 Nitrogen Only 10 85 Dung Only 4 109 Nitrogen&Dung 14 127

Using Minitab we first enter the data in table form.

Next, using the command STAT > TABLES > CHI SQUARE TEST, we select C2 and C3 and proceed to perform the Chi Square test for association.

```
Chi-Square Test: Blackleg, Clean

Expected counts are printed below observed counts

Blackleg    Clean    Total
1       16       85      101
9.88    91.12                  <- Expected counts = n(i,.)*n(.,j)/n(.,.)
2       10       85       95            n(i,.) is row i total (101,95,113 or 141)
9.29    85.71                     n(.,j) is column j total (44 or 406)
3        4      109      113            n(.,.) is overall total = 450
11.05   101.95
4       14      127      141
13.79   127.21
Total       44      406      450

Chi-Sq =  3.798 +  0.412 +
0.054 +  0.006 +
4.497 +  0.487 +
0.003 +  0.000 = 9.258   <- Chi Square test statistic (see Page 504).
DF = 3, P-Value = 0.026            <- p-value is less than Type I error rate suggesting we
reject the null hypothesis of independence and conclude
that the Presence of Blackleg is not independent of
fertilization.
```

Note that once we reject the null hypothesis of independence we start to look for where the dependency exists. If we compare the expected frequencies to the observed frequencies we find that we had more Blackleg with the NO fertilization treatment (obs: 16, exp: 10) than expected and less Blackleg with the Dung only fertilization than expected (obs:4, exp: 11). This suggests that any fertilizer will provide some protection for Blackleg but Dung might actually reduce the occurrance.

```
```