Question

S. tose question 2's results to help you solve) Suppose the class observed that there were 234 plants that were pink petal/spiked leaves and 42 that were pink petal/round leaves. Do you accept or reject your null hypothesis? Degrees of Freedom 2 4 5 6 7 8 0.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51 0.01一ー16.64一一9.32一11.34-13.28ー15.09一16.81一18.48-120.09ー! Pink + spiked - 0-15 Pink rounded = 0.25 am 1 Practice

Answer 1

Solution:-

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

Null hypothesis: The observed number of leaves are as per given distribution.

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

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 = 2 - 1
D.F = 1
(E_i) = n * p_i

$\chi ^{2}=\sum \left [\frac{(O_{r,c}-E_{r,c})^{2}}{E_{r,c}} \right ]$
X² = 14.09

X²_Critical = 3.84

Reject H₀, if X² > 3.84

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 1 degrees of freedom is more extreme than 14.09.

We use the Chi-Square Distribution Calculator to find P(X² > 14.09) = less than 0.001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.