Q 12. You say tomato The paper 鈥淟ink... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 11: Q 12. (page 718)

You say tomato The paper 鈥淟inkage Studies of the Tomato鈥� (Transactions of the Canadian Institute, $1931$ ) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio $9 : 3 : 3 : 1$
Assume that the conditions for inference are met. Carry out a test at the $伪 = 0.05$ significance level of the proposed genetic model.

Short Answer

Expert verified

$P -$ value is greater than the significance level which means that fail to reject $H_{0}$

Step by step solution

Given information

Significance level: $5 % = 0.05$

Ratio: $R_{1} : R_{2} : W_{1} : W_{2} = 9 : 3 : 3 : 1$

$n = 4$

Concept

Null hypothesis: H0

When the value of $P$ is less than or equal to the significance level, the null hypothesis is rejected.

Calculation

The null hypothesis: $H_{0}$

The phenotype occur in a ratio of $9 : 3 : 3 : 1$

$P_{t c} = \frac{9}{16}, P_{t p} = \frac{9}{16}, P_{d c} = \frac{3}{16}, P_{d p} = \frac{1}{16}$

Alternative hypothesis: H1

One or more of the offspring proportions are wrong.

Now,

Take the sum: $S = 9 + 3 + 3 + 1 = 16$

Create a table with the observed and predicted values:

The expected value is calculated as:

$E_{1} = T 脳 \frac{R_{1}}{R} = 1611 脳 \frac{9}{16} = 906.19 E_{2} = T 脳 \frac{R_{2}}{R} = 1611 脳 \frac{3}{16} = 302.06$

$E_{3} = T 脳 \frac{W_{1}}{R} = 1611 脳 \frac{3}{16} = 302.06 E_{4} = T 脳 \frac{W_{2}}{R} = 1611 脳 \frac{1}{16} = 100.69$

Now, use the test 鈭抯tatistics:

$x^{2} = {鈭�}_{i = 1}^{4} \frac{{(O_{i} 鈭� E_{i})}^{2}}{E_{i}} x^{2} = \frac{{(926 鈭� 906.19)}^{2}}{906.19} + \frac{{(288 鈭� 302.06)}^{2}}{302.06} + \frac{{(293 鈭� 302.06)}^{2}}{302.06} + \frac{{(104 鈭� 100.69)}^{2}}{100.69} x^{2} = 1.468$

Degree of freedom:

$d f = n 鈭� 1 = 4 鈭� 1 = 3$

For $x^{2} = 1.468$ and $d f = 3$ the $P -$ value is $0.6900$

So, the null hypothesis is not rejected because the $P -$ value is bigger than the significance level, implying that there is no difference in the distribution of generic features.

$P > 鈭� : 0.6900 > 0.05$ fail to reject $H_{0}$

As a result, there is insufficient evidence to demonstrate that the proposed generic model of tomato plant does not follow the supplied standard ratio.

Therefore, $P > 鈭� : 0.6900 > 0.05$