Question

An earlier statistics class used to have a lab. During lab students would count corn kernels. Here are some counts for corn kernels: purple round: 212 purple wrinkled: 87 yellow round: 85 yellow wrinkled: 19 Is there any reason to doubt the expected ration of 9:3:3:1 (the counts above are given in order)

Answer 1

Answer:

There is no reason to doubt the expected ratio of 9:3:3:1.

Step-by-step explanation:

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

$\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}$

The information provided is:

Counts for corn kernels:

Purple round: 212

Purple wrinkled: 87

Yellow round: 85

Yellow wrinkled: 19

TOTAL: 403

The expected ratio is, 9 : 3 : 3 : 1

Compute the expected count of corn kernels as follows:

E (Purple round) $=\frac{9}{9+3+3+1}\times 403=226.6875$

E (Purple wrinkled) $=\frac{3}{9+3+3+1}\times 403=75.5625$

E (Yellow round) $=\frac{3}{9+3+3+1}\times 403=75.5625$

E (Yellow wrinkled) $=\frac{1}{9+3+3+1}\times 403=25.1875$

Compute the test statistic as follows:

$\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}$

    $=[\frac{(212-226.6875)^{2}}{226.6875}]+[\frac{(87-75.5625)^{2}}{75.5625}]+[\frac{(85-75.5625)^{2}}{75.5625}]+[\frac{(19-25.1875)^{2}}{25.1875}]$

    $=5.382$

The test statistic value is, 5.382.

The degrees of freedom of the test is:

n - 1 = 4 - 1 = 3

Compute the p-value of the test as follows:

p-value = 0.1459

The p-value of the test is very large. So, the null hypothesis will not be rejected at any commonly used significance level.

Thus, concluding that the there is no reason to doubt the expected ratio of 9:3:3:1.