Friday, May 19, 2023

Tissue

 

Anatomy and Primary Structure of Monocot Stem-maize Stem



The outline of the maize in transverse section is more or less circular.

 

Primary Structure of Monocot Stem-maize Stem

 

The outline of the maize in transverse section is more or less circular. The transverse section of the monocot stem [maize] shows the following plan of arrangement of tissues from the periphery to the centre.

 Epidermis

 It is the outermost layer of the stem. It is made up of single layer of tightly packed parenchymatous cells. Their outer walls are covered with thick cuticle. The continuity of this layer may be broken here and there by the presence of a few stomata. There are no epidermal outgrowths.

 Hypodermis

 A few layer of sclerenchymatous cells lying below the epidermis constitute the hypodermis. This layer gives mechanical strength to the  plant. It is interrupted here and there by chlorenchyma cells.

 Ground Tissue

 There is no distinction into cortex, endodermis, pericycle and pith. The entire mass of parenchyma cells lying inner to the hypodermis forms the ground tissue.

The cell wall is made up of cellulose. The cells contain reserve food material like starch. The cells of the ground tissue next to the hypodermis are smaller in size, polygonal in shape and compactly arranged.

Towards the centre, the cells are loosely arranged, rounded in shape and bigger in size. The vascular bundles lie embedded in this tissue. The ground tissue stores food and performs gaseous exchange.

Vascular Bundles

Vascular bundles are scattered (atactostele) in the parenchymatous ground tissue. Each  vascular bundle is surrounded by a sheath of sclerenchy-matousfibres called bundle sheath. The vascular bundles areconjoint, collateral, endarch and closed.Vascular bundles are numerous, small and closely arranged in the peripheral portion. Towards the cen-tre, the bundles are comparatively large in size and loosely arranged­. Vascular bun-dles are skull or oval shaped.

 Phloem

The phloem in the monocot stem consists of sieve tubes and companion cells. Phloem parenchyma and phloem fibres are absent. It can be distinguished into an outer crushed protophloem and an inner metaphloem.

 Xylem

 Xylem vessels are arranged in the form of ’Y’ the two metaxylem vessels are located at the upper two arms and one or two protoxylem vessels at the base. In a mature bundle, the lowest protoxylem disintegrates and forms a cavity known as protoxylem lacuna.

 



Anatomical differences between dicot stem and monocot stem



1. Hypodermis

Characters of Dicot Stem :Collenchymatous

Characters of Monocot Stem :Sclerenchymatous

2. Ground tissue

Characters of Dicot Stem : Differentiated into cortex, endodermis and pericycle and pith

Characters of Monocot Stem : Not differentiated, but it is a continuous mass of parenchyma

3. Starch Sheath

Characters of Dicot Stem : Present

Characters of Monocot Stem : Absent

4. Medullary rays

Characters of Dicot Stem : Present

Characters of Monocot Stem : Absent

5. Vascular bundles

Characters of Dicot Stem : (a) Collateral and open

Characters of Monocot Stem : (a) Collateral and closed

 Characters of Dicot Stem : (b) Arranged in a ring

Characters of Monocot Stem : (b) Scattered in ground tissue

 Characters of Dicot Stem : (c) Secondary growth occurs

Characters of Monocot Stem : (c) Secondary growth usually does not occur.

Anatomical differences between root and stem

  

 1. Epidermis 

Root : Absence of cuticle and epidermal pores.

Stem : Presence of cuticle and epidermal pores.

Root : Presence of unicellular root hairs.

Stem : Presence of unicellular and multicellular trichomes

2. Outer Cortical cells

Root :Chlorenchyma absent 

Stem :Chlorenchyma present

3. Endodermis 

Root : Well defined 

Stem : ill-defined or absent.

4. Vascular bundles

Root : Radial arrangement 

Stem : Conjoint arrangement

5. Xylem 

Root : Exarch 

Stem :Endarch


Dicot Root


Monocot Root




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Wednesday, May 17, 2023

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Monday, May 15, 2023

Anova Test

ANOVA Test - Types, Table, Formula, Examples

ANOVA Test

ANOVA Test is used to analyze the differences among the means of various groups using certain estimation procedures. ANOVA means analysis of variance. ANOVA test is a statistical significance test that is used to check whether the null hypothesis can be rejected or not during hypothesis testing.

An ANOVA test can be either one-way or two-way depending upon the number of independent variables. In this article, we will learn more about an ANOVA test, the one-way ANOVA and two-way ANOVA, its formulas and see certain associated examples.

1. What is ANOVA Test?
2. ANOVA Formula
3. One Way ANOVA
4. Two Way ANOVA
5. FAQs on ANOVA Test

What is ANOVA Test?

ANOVA test, in its simplest form, is used to check whether the means of three or more populations are equal or not. The ANOVA test applies when there are more than two independent groups. The goal of the ANOVA test is to check for variability within the groups as well as the variability among the groups. The ANOVA test statistic is given by the f test .

ANOVA Test Definition

ANOVA test can be defined as a type of test used in hypothesis testing to compare whether the means of two or more groups are equal or not. This test is used to check if the null hypothesis can be rejected or not depending upon the statistical significance exhibited by the parameters. The decision is made by comparing the ANOVA test statistic with the critical value.

ANOVA Test Example

Suppose it needs to be determined if consumption of a certain type of tea will result in a mean weight loss. Let there be three groups using three types of tea - green tea, earl grey tea, and jasmine tea. Thus, to compare if there was any mean weight loss exhibited by a certain group, the ANOVA test (one way) will be used.

Suppose a survey was conducted to check if there is an interaction between income and gender with anxiety level at job interviews. To conduct such a test a two-way ANOVA will be used.

ANOVA Formula

ANOVA Table

There are several components to the ANOVA formula. The best way to solve a problem on an ANOVA test is by organizing the formulas into an ANOVA table. The ANOVA formulas are given below.

Sum of squares between groups, SSB = \(\sum n_{j}(\overline{X}_{j}-\overline{X})^{2}\). Here, \(\overline{X}_{j}\) is the mean of the jth group, \(\overline{X}\) is the overall mean and \(n_{j}\) is the sample size of the jth group.

\(\overline{X}\) = \(\frac{\overline{X}_{1} + \overline{X}_{2} + \overline{X}_{3} + ... + \overline{X}_{j}}{j}\)

Sum of squares of errors, SSE = \(\sum\sum(X-\overline{X}_{j})^{2}\). Here, X refers to each data point in the jth group.

Total sum of squares, SST = SSB + SSE

Degrees of freedom between groups, df1 = k - 1. Here, k denotes the number of groups.

Degrees of freedom of errors, df2 = N - k, where N denotes the total number of observations across k groups.

Total degrees of freedom, df3 = N - 1.

Mean squares between groups, MSB = SSB / (k - 1)

Mean squares of errors, MSE = SSE / (N - k)

ANOVA test statistic, f = MSB / MSE

Critical Value at \(\alpha\) = F(\(\alpha\), k - 1, N - k)

ANOVA Table

The ANOVA formulas can be arranged systematically in the form of a table. This ANOVA table can be summarized as follows:

Source of Variation Sum of Squares Degrees of Freedom Mean Squares F Value
Between Groups SSB = Σnj (\(\overline{X}_{j}-\overline{X})^{2}\) df1 = k - 1 MSB = SSB / (k - 1) f = MSB / MSE
Error SSE = ΣΣ(\(X-\overline{X}_{j})^{2}\) df2 = N - k MSE = SSE / (N - k)  
Total

SST = SSB + SSE

 df3 = N - 1    

One Way ANOVA

The one way ANOVA test is used to determine whether there is any difference between the means of three or more groups. A one way ANOVA will have only one independent variable. The hypothesis for a one way ANOVA test can be set up as follows:

Null Hypothesis, \(H_{0}\): \(\mu_{1}\) = \(\mu_{2}\) = \(\mu_{3}\) = ... = \(\mu_{k}\)

Alternative Hypothesis, \(H_{1}\): The means are not equal

Decision Rule: If test statistic >critical value then reject the null hypothesis and conclude that the means of at least two groups are statistically significant.

The steps to perform the one way ANOVA test are given below:

  • Step 1: Calculate the mean for each group.
  • Step 2: Calculate the total mean. This is done by adding all the means and dividing it by the total number of means.
  • Step 3: Calculate the SSB.
  • Step 4: Calculate the between groups degrees of freedom.
  • Step 5: Calculate the SSE.
  • Step 6: Calculate the degrees of freedom of errors.
  • Step 7: Determine the MSB and the MSE.
  • Step 8: Find the f test statistic.
  • Step 9: Using the f table for the specified level of significance, \(\alpha\), find the critical value. This is given by F(\(\alpha\), df1 . df2 ).
  • Step 10: If f >F then reject the null hypothesis.

Limitations of One Way ANOVA Test

The one way ANOVA is an omnibus test statistic. This implies that the test will determine whether the means of the various groups are statistically significant or not. However, it cannot distinguish the specific groups that have a statistically significant mean. Thus, to find the specific group with a different mean, a post hoc test needs to be conducted.

Two Way ANOVA

The two way ANOVA has two independent variables. Thus, it can be thought of as an extension of a one way ANOVA where only one variable affects the dependent variable. A two way ANOVA test is used to check the main effect of each independent variable and to see if there is an interaction effect between them. To examine the main effect, each factor is considered separately as done in a one way ANOVA. Furthermore, to check the interaction effect, all factors are considered at the same time. There are certain assumptions made for a two way ANOVA test. These are given as follows:

  • The samples drawn from the population must be independent.
  • The population should be approximately normally distributed.
  • The groups should have the same sample size.
  • The population variances are equal

Suppose in the two way ANOVA example, as mentioned above, the income groups are low, middle, high. The gender groups are female, male, and transgender. Then there will be 9 treatment groups and the three hypotheses can be set up as follows:

\(H_{01}\): All income groups have equal mean anxiety.

\(H_{11}\): All income groups do not have equal mean anxiety.

\(H_{02}\): All gender groups have equal mean anxiety.

\(H_{12}\): All gender groups do not have equal mean anxiety.

\(H_{03}\): Interaction effect does not exist

\(H_{13}\): Interaction effect exists.

  • ANOVA test is used to check whether the means of three or more groups are different or not by using estimation parameters such as the variance.
  • An ANOVA table is used to summarize the results of an ANOVA test.
  • There are two types of ANOVA tests - one way ANOVA and two way ANOVA
  • One way ANOVA has only one independent variable while a two way ANOVA has two independent variables.
 

Examples on ANOVA Test

  1. Example 1: Three types of fertilizers are used on three groups of plants for 5 weeks. We want to check if there is a difference in the mean growth of each group. Using the data given below apply a one way ANOVA test at 0.05 significant level.

    Fertilizer 1 Fertilizer 2 Fertilizer 3
    6 8 13
    8 12 9
    4 9 11
    5 11 8
    3 6 7
    4 8 12

    Solution:

    \(H_{0}\): \(\mu_{1}\) = \(\mu_{2}\) = \(\mu_{3}\)

    \(H_{1}\): The means are not equal

    Fertilizer 1 Fertilizer 2 Fertilizer 3
    6 8 13
    8 12 9
    4 9 11
    5 11 8
    3 6 7
    4 8 12
    \(\overline{X}_{1}\) = 5 \(\overline{X}_{1}\) = 9 \(\overline{X}_{1}\) = 10

    Total mean, \(\overline{X}\) = 8

    \(n_{1}\) = \(n_{2}\) = \(n_{3}\) = 6, k = 3

    SSB = 6(5 - 8)2 + 6(9 - 8)2 + 6(10 - 8)2

    = 84

    df1 = k - 1 = 2

    Fertilizer 1 (X - 5)2 Fertilizer 2 (X - 9)2 Fertilizer 3 (X - 10)2
    6 1 8 1 13 9
    8 9 12 9 9 1
    4 1 9 0 11 1
    5 0 11 4 8 4
    3 4 6 9 7 9
    4 1 8 1 12 4
    \(\overline{X}_{1}\) = 5 Total = 16 \(\overline{X}_{1}\) = 9 Total = 24 \(\overline{X}_{1}\) = 10 Total = 28

    SSE = 16 + 24 + 28 = 68

    N = 18

    df2 = N - k = 18 - 3 = 15

    MSB = SSB / df1 = 84 / 2 = 42

    MSE = SSE / df2 = 68 / 15 = 4.53

    ANOVA test statistic, f = MSB / MSE = 42 / 4.53 = 9.33

    Using the f table at \(\alpha\) = 0.05 the critical value is given as F(0.05, 2, 15) = 3.68

    As f >F, thus, the null hypothesis is rejected and it can be concluded that there is a difference in the mean growth of the plants.

    Answer: Reject the null hypothesis

  2. Example 2: A trial was run to check the effects of different diets. Positive numbers indicate weight loss and negative numbers indicate weight gain. Check if there is an average difference in the weight of people following different diets using an ANOVA Table.

    Low Fat Low Calorie Low Protein Low Carbohydrate
    8 2 3 2
    9 4 5 2
    6 3 4 -1
    7 5 2 0
    3 1 3 3

    Solution:

    \(H_{0}\): \(\mu_{1}\) = \(\mu_{2}\) = \(\mu_{3}\) = \(\mu_{4}\)

    \(H_{1}\): The means are not equal

    Low Fat (X - 6.6)2 Low Calorie (X - 3)2 Low Protein (X - 3.4)2 Low Carbohydrate (X - 1.2)2
    8 2 2 1 3 0.2 2 0.6
    9 5.8 4 1 5 2.6 2 0.6
    6 0.4 3 0 4 0.4 -1 4.8
    7 0.2 5 4 2 2 0 1.4
    3 13 1 4 3 0.2 3 3.2
    \(\overline{X}_{1}\) = 6.6 Total = 21.4 \(\overline{X}_{2}\) = 3 Total = 10 \(\overline{X}_{3}\) = 3.4 Total = 5.4 \(\overline{X}_{4}\) = 1.2 Total = 10.6

    Total mean, \(\overline{X}\) = 3.6

    \(n_{1}\) = \(n_{2}\) = \(n_{3}\) = \(n_{4}\) = 5, k = 4

    SSB = \(n_{1}(\overline{X}_{1}-\overline{X})^{2}\) + \(n_{2}(\overline{X}_{2}-\overline{X})^{2}\) +&\(n_{3}(\overline{X}_{3}-\overline{X})^{2}\) +\(n_{4}(\overline{X}_{4}-\overline{X})^{2}\)

    = 75.8

    SSE = 21.4 + 10 + 5.4 + 10.6 = 47.4

    The ANOVA Table can be constructed as follows:

    Source of Variation Sum of Squares Degrees of Freedom Mean Squares F Value
    Between Groups SSB = Σnj (\(\overline{X}_{j}-3.6)^{2}\)
    = 75.8     		df1 = k - 1
    = 4 - 1
    = 3     		MSB = SSB / (k - 1)
    = 25.3     		f = MSB / MSE
    = 8.43
    Error SSE = ΣΣ(\(X-\overline{X}_{j})^{2}\)
    = 47.4     		df2 = N - k
    = 20 - 4
    = 16     		MSE = SSE / (N - k)
    = 3     		 
    Total

    SST = SSB + SSE
    = 123.2

    		df3 = N - 1
    = 19     		   

    As no significance level is specified, \(\alpha\) = 0.05 is chosen.

    F(0.05, 3, 16) = 3.24

    As 8.43 >3.24, thus, the null hypothesis is rejected and it can be concluded that there is a mean weight loss in the diets.

    Answer: Reject the null hypothesis

  3. Example 3: Determine if there is a difference in the mean daily calcium intake for people with normal bone density, osteopenia, and osteoporosis at a 0.05 alpha level. The data was recorded as follows:

    Normal Density Osteopenia Osteoporosis
    1200 1000 890
    1000 1100 650
    980 700 1100
    900 800 900
    750 500 400
    800 700 350

    Solution:

    Using the ANOVA test the hypothesis is set up as follows:

    \(H_{0}\): \(\mu_{1}\) = \(\mu_{2}\) = \(\mu_{3}\)

    \(H_{1}\): The means are not equal

    Normal Density (X - 938.3)2 Osteopenia (X - 800)2 Osteoporosis (X - 715)2
    1200 68,486.9 1000 40,000 890 30,625
    1000 3,806.9 1100 90,000 650 4,225
    980 1,738.9 700 10,000 1100 148,225
    900 1,466.9 800 0 900 34,225
    750 35,456.9 500 90,000 400 99,225
    800 19,126.9 700 10,000 350 133,225
    \(\overline{X}_{1}\) = 938.3 Total = 130,083.3 \(\overline{X}_{2}\) = 800 Total = 240,000 \(\overline{X}_{3}\) = 715 Total = 449,750

    Total mean, \(\overline{X}\) = 817.8

    \(n_{1}\) = \(n_{2}\) = \(n_{3}\) = 6, k = 3

    SSB = \(n_{1}(\overline{X}_{1}-\overline{X})^{2}\) + \(n_{2}(\overline{X}_{2}-\overline{X})^{2}\) + \(n_{3}(\overline{X}_{3}-\overline{X})^{2}\)

    = 152,477.7

    SSE = 130,083.3 + 240,000 + 449,750 = 819,833.3

    The ANOVA Table can be constructed as follows:

    Source of Variation Sum of Squares Degrees of Freedom Mean Squares F Value
    Between Groups SSB = Σnj (\(\overline{X}_{j}-817.8)^{2}\)
    = 152,477.7     		df1 = k - 1
    = 3 - 1
    = 2     		MSB = SSB / (k - 1)
    = 76,238.6     		f = MSB / MSE
    = 1.395
    Error SSE = ΣΣ(\(X-\overline{X}_{j})^{2}\)
    = 819,833.3     		df2 = N - k
    = 18 - 3
    = 15     		MSE = SSE / (N - k)
    = 54,655.5     		 
    Total

    SST = SSB + SSE
    = 972,311

    		df3 = N - 1
    = 17     		   

    Using the F table the critical value is F(0.05, 2, 15) = 3.68

    As 1.395 <3.68, the null hypothesis cannot be rejected and it is concluded that there is not enough evidence to prove that the mean daily calcium intake of the three groups is different.

    Answer: Do not reject the null hypothesis

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