Question

Seedlings of the parasitic plant Cuscuta pentagona (dodder) hunt by directing growth preferentially toward nearby host plants. To investigate the possibility that the parasite detects volatile chemicals produced by host plants, a researcher placed individual dodder seedlings into a vial of water at the center of a circular paper disc. A chamber containing volatile extracts from tomato (a host plant) was placed at one edge of the disc, whereas a control chamber containing only solvent was placed at the opposite end. The researcher divided the disk in to 4, equal-area quadrats to record which direction the seedlings grew. 30 dodder plants were tested, and 17 grew toward the volatiles, 2 grew away from the volatiles (toward the solvent), 7 grew toward the left quadrant, and 4 grew toward the right quadrant.

1) Is this an experimental or observational study?

2) State null and alternative hypothesis

3) Which test is appropriate for this study? T-test, contingency test or goodness of fit?

Answer 1

Solution:-

1) Experimental study.

2)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The proportion of grown seedling are same for the four parts in the disk.

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

3) Goodness of fit.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 4 - 1
D.F = 3
(E_i) = n * p_i


X² = 17.733

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 17.733.

We use the Chi-Square Distribution Calculator to find P(X² > 17.733) = 0.001.

Interpret results. Since the P-value (0.001) is less than the significance level (0.05), we cannot accept the null hypothesis.