Monthly Archives: August 2014

Beat the IELTS Task 2 Writing

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The task 2 writing on the IELTS calls on students to express their opinion about a topic. This is not as easy as it sounds even for native speakers. There are many common pitfalls such as not responding to the question or not understanding what the question wants you to do. One of the first steps to take in writing a response to task 2 question is to break down the question to determine what you need to do.

Many Task 2 writing prompts have three components to them. They are listed below

  1. A statement that is a fact or opinion
  2. What you need to do (the job)
  3. Advice on how to complete the task

Let’s look at an example

  • Schools should ask students to evaluate their teachers. Do you agree or disagree? Use specific reasons and examples to support your answer.

In this example, we have all three components.

  1. Schools should ask students to evaluate their teachers. (This is the opinion you are reacting to)
  2. Do you agree or disagree? (Your job is to explain why you agree or disagree)
  3. Use specific reasons and examples to support your answer. (Here is the advice to complete the task)

In this example, our job is to agree or disagree about whether students should evaluate teachers. This example is a one job task. In other words, you have to only do one thing which explains why you agree or disagree. Some writing prompts call for doing more than one job such as compare and contrast in which you compare and then you contrast. One job tasks are the easiest to respond to.

Another important point is that if the prompt asks you to agree or disagree this is what you should do. It is too complicated to try and agree and disagree because it takes a much higher level of English to express a nuance opinion. Keep it simple and maximize your score through simply agreeing or disagreeing. Everybody knows the world is more complicated then that but if you need to take the IELTS you might not be ready to express this yet. Don’t try to show the reader how smart you are save that for the future.

Outline
The biggest mistake many students make is they jump right in to writing without developing any sort of outline. This is similar to jumping in your car to drive somewhere you have never been without directions. You’ll eventually get there but you journey is longer and unpredictable because of lack of preparation. It is important to make a simple outline of what you want to say.

Below is one way to approach a one job Task 2 writing prompt. It uses a traditional 5 paragraph essay format.

  1. Introduction-Paragraph has three component to it as explained below
    1. The 1st sentence should restate the statement or opinion. Indicate what the topic is even though the reader already knows.
    2. Indicate whether you agree or disagree. Tell the reader if you agree or disagree right away. There is no time to be indirect and mysterious
    3. Give your reasons for agreeing or disagreeing. In order to have five paragraphs you will have to develop three reasons why you agree or disagree. Each reason will have its own paragraph in which you explain it. It is common for students to struggle here. They have an opinion but they do not have any well thought out reason for the opinon. This is one reason why the IELTS is not only an English test but a test of thinking ability.
  2. Body paragraph-The three body paragraphs follow the same format as explained below
    1. 1st sentence should state the reason-Your first sentence in each body paragraph should restate one of your reasons why you agree or disagree.
    2. Example-Every reason needs some sort of illustration that further explains the reason. For example, if you think smoking is bad for someone because it causes health problems. You might share a story about how smoking killed a close relative. This illustration further clarifies why you think smoking is bad for you
  3.  Conclusion-Restate your opinion and reasons using different English if possible. There are other ways to end an essay but this is the simplest.

Examples will be provided in the future

Chi-Square Goodness-of-Fit-Test

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The chi-square test is a non-parametric test that is used in statistic to determine if an observed distribution or model conforms or is similar to an expected distribution or model. In simple terms, this test will tell you if the data you collected is similar to other data or to what you expected.

There are several types of chi-square test such as the Chi-square Test of Independence, which is used for nominal data, and the Goodness-of-Fit Test, which deals with data that is not nominal. This post is about the Goodness-of-Fit Test. The Goodness-of-Fit test compares the distribution of the observed data with an expected distribution.

A unique caveat of chi-square test is that we normally desire as a researcher to make sure we do not reject our model. This is opposite of traditional hypothesis testing which desires often to reject the null hypothesis as this indicates that there is a statistical difference. With chi-square test, we want our observed model to be similar to the values found in the expected model. What this means is that our model represents what is happening in the real-world and is not only theoretical. If we reject the null it means that the model we are trying to create is not similar to expected values that might be found in the real world. In other words, we found something that does not conform to what is expected. If a model does not represent the world, it may not serve much purpose.

Here are the assumptions of Goodness-of-Fit Test

  • Random selection of subjects
  • Mutually exclusive categories

Here are the steps

  1. Determine hypothesis
    • H0: There is no difference between the observed values/model and the expected values/model
    • H1: There is a difference between the observed values/model and the expected values/model
  2. Decide level of significance
  3. Determine degree of freedom to find chi-square critical
  4. Compute for the expected frequencies
  5. Compute chi-square
  6. Make decision to accept or reject null
  7. State conclusion

Here is an example

A principal wants to know if the number of students absent each day of the week is the same. Below are the results for one week.

Day                  Absents

Monday                 17

Tuesday                 20

Wednesday            16

Thursday               14

Friday                    13

Step 1: Determine Hypothesis

  • H0: The number of students absent is the same every day
  • H1: The number of students absent is not the same every day

Step 2: Decide level of significance

  • 0.05

Step 3 Determine chi-square critical region (computer does this for you)

  • Chi-square critical region = 9.48

Step 4: Compute expected frequencies

  • Computer does this

Step 5: Compute Chi square (computer does this for you)

  • Chi-square = 1.87

Step 6: Make decision

  • Since the computed chi-square of 1,87 is less than the critical chi-square value of 9.48 we do not reject the null hypothesis

Step 7: Conclusion

  • Since we do not reject the null hypothesis we can say that there is a lack of evidence that there is a difference in the number of absences each day of the week. In other words, the number of students absent each day is the same.

NOTE: There is also a way to do this test when the expected frequencies are unequal

Simple Linear Regression Analysis

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Simple linear regression analysis is a technique that is used to model the dependency of one dependent variable upon one independent variable. This relationship between these two variables is explained by an equation.

When regression is employed normally the data points are graphed on a scatterplot. Next, the computer draws what is called the “best-fitting” line. The line is the best fit because it reduces the amount of error between actual values and predicted values in the model. The official name of the model is the least square model in that it is the model with the least amount of error. As such, it is the best model for predicting future values

It is important to remember that one of the great enemies of statistics is explaining error or residual. In general, any particular data point that is not the mean is said to have some error in it. For example, if the average is 5 and one of the data points is three 5 -3 = 2 or an error of 2. Statistics often want to explain this error. What is causing this variation from the mean is a common question.

There are two ways that simple regression deals with error

  1. The error cannot be explained. This is known as unexplained variation.
  2. The error can be explained. This is known as explained variation.

When these two values are added together you get the total variation which is also known as the “sum of squares for error.”

Another important term to be familiar with is the standard error of estimate. The standard error of estimate is a measurement of the standard deviation of the observed dependent variables values from predicted values of the dependent variable. Remember that there is always a slight difference between observed and predicted values and the model wants to explain as much of this as possible.

In general, the smaller the standard error the better because this indicates that there is not much difference between observed data points and predicted data points. In other words, the model fits the data very well.

Another name for the explained variation is the coefficient of determination. The coefficient of determination is the amount of variation that is explained by the regression line and the independent variable. Another name for this value is the r². The coefficient of determination is standardized to have a value between 0 to 1 or 0% to 100%.

The higher your r² the better your model is at explaining the dependent variable. However, there are a lot of qualifiers to this statement that goes beyond this post.

Here are the assumptions of simple regression

  • Linearity–The mean of each error is zero
  • Independence of error terms–The errors are independent of each other
  • Normality of error terms–The error of each variable is normally distributed
  • Homoscedasticity–The variance of the error for the value of each variable is the same

There are many ways to check all of this in SPSS which is beyond this post.

Below is an example of simple regression using data from a previous post

You want to know how strong is the relationship of the exam grade on the number of words in the students’ essay. The data is below

Student         Grade        Words on Essay
1                             79                           147
2                             76                           143
3                             78                           147
4                             84                           168
5                             90                           206
6                             83                           155
7                             93                           192
8                             94                           211
9                             97                           209
10                          85                           187
11                          88                           200
12                          82                           150

Step 1: Find the Slope (The computer does this for you)
slope = 3.74

Step 2: Find the mean of X (exam grade) and Y (words on the essay) (Computer does this for you)
X (Exam grade) = 85.75        Y (Words on Essay) = 176.25

Step 3: Compute the intercept of the simple linear regression (computer does this)
-145.27

Step 4: Create linear regression equation (you do this)
Y (words on essay) = 3.74*(exam grade) – 145.27
NOTE: you can use this equation to predict the number of words on the essay if you know the exam grade or to predict the exam grade if you know how many words they wrote in the essay. It is simple algebra.

Step 5: Calculate Coefficient of Determination r² (computer does this for you)
r² = 0.85
The coefficient of determination explains 85% of the variation in the number of words on the essay. In other words, exam grades strongly predict how many words a student will write in their essay.

Spearman Rank Correlation

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Spearman rank correlation aka ρ is used to measure the strength of the relationship between two variables. You may be already wondering what is the difference between Spearman rank correlation and Person product moment correlation. The difference is that Spearman rank correlation is a non-parametric test while Person product moment correlation is a parametric test.

A non-parametric test does not have to comply with the assumptions of parametric test such as the data being normally distributed. This allows a researcher to still make inferences from data that may not have normality. In addition, non-parametric test are used for data that is at the ordinal or nominal level. In many ways, Spearman correlation and Pearson product moment correlation compliment each other. One is used in non-parametric statistics and the other for parametric statistics and each analyzes the relationship between variables.

If you get suspicious results from your Pearson product moment correlation analysis or your data lacks normality Spearman rank correlation may be useful for you if you still want to determine if there is a relationship between the variables. Spearmen correlation works by ranking the data within each variable. Next, the Pearson product moment correlation is calculated between the two sets of rank variables. Below are the assumptions of Spearman correlation test.

  • Subjects are randomly selected
  • Observations are at the ordinal level at least

Below are the steps of Spearman correlation

  1. Setup the hypotheses
    1. H0: There is no correlation between the variables
    2. H1: There is a correlation between the variables
  2. Set the level of significance
  3. Calculate the degrees of freedom and find the t-critical value (computer does this for you)
  4. Calculate the value of Spearman correlation or ρ (computer does this for you)
  5. Calculate the t-value(computer does this for you) and make a statistical decision
  6. State conclusion

Here is an example

A clerk wants to see if there is a correlation between the overall grade students get on an exam and  the number of words they wrote for their essay. Below are the results

Student         Grade        Words on Essay
1                             79                           147
2                             76                           143
3                             78                           147
4                             84                           168
5                             90                           206
6                             83                           155
7                             93                           192
8                             94                           211
9                             97                           209
10                           85                           187
11                           88                           200
12                           82                           150

Note: The computer will rank the data of each variable with a rank of 1 being the highest value of a variable and a rank 12 being the lowest value of a variable. Remember that the computer does this for you.

Step 1: State hypotheses
H0: There is no relationship between grades and words on the essay
H1: There is a relationship between grades and words on the essay

Step 2: Determine level of significance
Level set to 0.05

Step 3: Determine critical t-value
t = + 2.228 (computer does this for you)

Step 4: Compute Spearman correlation
ρ = 0.97 (computer does this for you)
Note: This correlation is very strong. Remember the strongest relationship possible is + 1

Step 5: Calculate t-value and make a decision
t = 12.62   ( the computer does this for you)
Since the computed t-value of 12.62 is greater than the t-critical value of 2.228 we reject the null hypothesis

Step 6: Conclusion
Since the null hypotheses are rejected, we can conclude that there is evidence that there is a strong relationship between exam grade and the number of words written on an essay. This means that a teacher could tell students they should write longer essays if they want a higher grade on exams

Correlation

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A correlation is a statistical method used to determine if a relationship exists between variables.  If there is a relationship between the variables it indicates a departure from independence. In other words, the higher the correlation the stronger the relationship and thus the more the variables have in common at least on the surface.

There are four common types of relationships between variables there are the following

  1. positive-Both variables increase or decrease in value
  2. Negative- One variable decreases in value while another increases.
  3. Non-linear-Both variables move together for a time then one decreases while the other continues to increase
  4. Zero-No relationship

The most common way to measure the correlation between variables is the Pearson product-moment correlation aka correlation coefficient aka r.  Correlations are usually measured on a standardized scale that ranges from -1 to +1. The value of the number, whether positive or negative, indicates the strength of the relationship.

The Person Product Moment Correlation test confirms if the r is statistically significant or if such a relationship would exist in the population and not just the sample. Below are the assumptions

  • Subjects are randomly selected
  • Both populations are normally distributed

Here is the process for finding the r.

  1. Determine hypotheses
    • H0: r = 0 (There is no relationship between the variables in the population)
    • H0: r ≠ 0 (There is a relationship between the variables in the population)
  2. Decided what the level of significance will be
  3. Calculate degrees of freedom to determine the t critical value (computer does this)
  4. Calculate Pearson’s r (computer does this)
  5. Calculate t value (computer does this)
  6. State conclusion.

Below is an example

A clerk wants to see if there is a correlation between the overall grade students get on an exam and the number of words they wrote for their essay. Below are the results

Student         Grade        Words on Essay
1                             79                           147
2                             76                           143
3                             78                           147
4                             84                           168
5                             90                           206
6                             83                           155
7                             93                           192
8                             94                           211
9                             97                           209
10                          85                           187
11                          88                           200
12                          82                           150

Step 1: State Hypotheses
H0: There is no relationship between grade and the number of words on the essay
H1: There is a relationship between grade and the number of words on the essay

Step 2: Level of significance
Set to 0.05

Step 3: Determine degrees of freedom and t critical value
t-critical = + 2.228 (This info is found in a chart in the back of most stat books)

Step 4: Compute r
r = 0.93                       (calculated by the computer)

Step 5: Decision rule. Calculate t-value for the r

t-value for r = 8.00  (Computer found this)

Since the computed t-value of 8.00 is greater than the t-critical value of 2.228 we reject the null hypothesis.

Step 6: Conclusion
Since the null hypothesis was rejected, we conclude that there is evidence that a strong relationship between the overall grade on the exam and the number of words written for the essay. To make this practical, the teacher could tell the students to write longer essays if they want a better score on the test.

IMPORTANT NOTE

When a null hypothesis is rejected there are several possible relationships between the variables.

  • Direct cause and effect
  • The relationship between X and Y may be due to the influence of a third variable not in the model
  • This could be a chance relationship. For example, foot size and vocabulary. Older people have bigger feet and also a larger vocabulary. Thus it is a nonsense relationship

Two-Way Analysis of Variance

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Two-way analysis of variance is used when we want to know the following pieces of information.
• The means of the blocks or subpopulations
• The means of the treatment groups
• The means of the interaction of the subpopulation and treatment groups

Now you are probably confused but remember that two-way analysis of variance is an extension of randomized block designed. With randomized block design, there were two hypotheses one for the treatment groups and one for the blocks or subpopulations. What we are doing for two-analysis is assessing the interaction effect, which is the amount of the variation of
subpopulation and treatment group). The assessment of the interaction effect gives us the third hypothesis. To put it in simple words when both the subpopulation and the treatment are present combined they have some sort of influence just as they do when one or the other is present. Therefore, two-way analysis of variance is randomized block designed plus an interaction effect hypothesis.

Another important difference is the use of repeated measures. In a two-way analysis of variance, at least one of the groups received the treatment more than once. In a randomized block design, each group receives the treatment only one time. Your research questions determine if any group needs to experience the treatment more than once.

Below are the assumptions
• Sample randomly selected
• Populations have homogeneous standard deviations
• Population distributions are normal
• Population covariances are equal.

Here are the steps
1. Set up hypotheses (there will be three of them)
a.Treatment means (AKA factor A)
i. H0: There is no difference in the treatment means
ii. H1: H0 is false
b. Block means (AKA factor B)
i. H0: There is no difference in the block means
ii. H1: is false
c. Interaction between Factor A and B
i. H0: There is no interacting effect between factor A & B
ii. H1: There is an interacting effect between factor A & B
2. Determine your level of statistical significance
3. Determine F critical (there will be three now and the computer does this)
4. Calculate the F-test values (there will be three now and the computer does this)
5. Test hypotheses
6. State conclusion

Here is an example
A music teacher wants to study the effect of instrument type and service center on the repair time measured in minutes. Four instruments (sax, trumpet, clarinet, flute) were picked for the analysis. Each service center was assigned to perform the particular repair on two instruments in each category

Instrument
Service centers Sax Trumpet Clarinet Flute
1                        60      50          58         60
70      56          62         64
2                        50      53          48         54
54      57          64         46
3                        62      54          46         51
64      66          52         49

Here are your research questions
• Is there a difference in the means of the repair time between service centers?
• Is there a difference in the means of the repair time between instrument type?
• Is there an interaction due to service center and type of instrument on the mean of the repair time
Let us go through each of our steps
Step 1: State the hypotheses
• Treatment means (AKA factor A)
a. H0: There is no difference in the means of the service centers
b. H1: H0 is false
• Block means (AKA factor B)
a. H0: There is no difference in means of the instrument types
b. H1: is false
• Interaction between Factor A and B
a. H0: There is no interacting effect between service center and instrument type
b. H1: There is an interacting effect between service center and instrument type

Step 2: Significance level
• Set at 0.1

Step 3: Determine F-Critical
For the instruments, F-critical is 2.81
For the service centers, F-critical is 2.61
For the interaction effect, F-critical is 2.33

Step 4: Calculate F-values
Service centers 3.2
Instrument type 1.4
Interaction 2.1

Step 5: Make decision
Since the F-value of 3.2 is greater than the F-critical of 2.8 we reject the null hypothesis for the service centers

Since the F-value of 1.4 is less than the F-critical of 2.61 we do not reject the null hypothesis for the instrument types

Since the F-value of 2.1 is less than the F-critical of 2.3 we do reject the null hypothesis for the interaction effect of service center and instrument type.

Step 6: Conclusion
Since we reject the null hypothesis that there is no difference in the means of the repair time of the service centers, we conclude that there is evidence of a difference in the repair times between service centers. This means that one service center is faster than the others are. To find out, do a posthoc test.

Since we do not reject the null hypothesis that there is no difference in the means of the repair time of the instrument types, we conclude that there is no evidence of a difference in the repair time between instrument types. In other words, it does not matter what type of instrument is being fixed as they will all take about the same amount of time.

Since we do not reject the null hypothesis that there is no interaction effect of service center and instrument type on the mean of the repair time, we conclude that there is no evidence of an interaction effect of service center and instrument type on repair time. In other words, if service center and instrument type are considered at the same time there is no difference in how fast the instruments are repaired.

Analysis of Variance: Randomized Block Design

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Randomized blocked design is used when a researcher wants to compare treatment means. What is unique to this research design is that the experiment is divided into two or more mini-experiments.

The reason behind this is to reduce the variation within-treatments so that it is easier to find differences between means.  Another unique characteristic of randomized block design is that since there is more than one experiment happening at the same time, there will be more than one set of hypotheses to consider. There will be a set of hypotheses for the treatment groups and also for the block groups. The block groups are the several subpopulations with the sample. Below are the assumptions

  • Samples are randomly selected
  • Populations are homogeneous
  • Populations are normally distributed
  • Populations covariances are equal
    •  Covariance is a measure of the commonality that two variables deviate from their expected values. If two variable deviates in similar ways the covariance will be high and vice versa. The standardized version of covariance is correlation.

Looking at equations and doing this by hand is tough. It is better to use SPSS or excel to calculate results. We are going to look at an example and see an application of randomized block design.

A professor wants to see if “time of day” affects his students score on a quiz. He randomly divides his stat class into five groups and has them take the quiz at one of four times during the day.  Below are the results
Time Period/Treatment
Section    8-9                10-11                11-12                1-2
1                  25                      22                        20                     25
2                  28                      24                        29                     23
3                  30                      25                        25                     27
4                  24                      27                        28                     25
5                  21                      28                        30                     24

The treatment groups here are the time periods. The are along the time and are 8-9, 10-11, 11-12, 1-2. The block groups are along the left-hand side and the are section 1, 2, 3, 4, 5. The block groups are the 5 different experimental groups of the larger population of the statistics class. What is happening here is that all members from all groups all took the quiz at one of the four times. For example, members from section one took the quiz at 8-9, 10-11, 11-12, and 1-2. The same for group 2 and so forth.  By having five different groups take the quiz at each of the time periods it should hopefully improve the accuracy of the results. It is like sampling a population five times instead of one time.

In addition, by having four different time periods, we can hopefully see much more clearly if the time period makes a difference. We have four different time periods instead of two or three. Below are the steps for solving this problem.

Step 1: State hypotheses
For Time periods
Null hypothesis: There is no difference in the means between time periods
Alternative hypothesis: There is a difference in the means between time periods
For Blocks
Null hypothesis: There is no difference in the means among the sections of students
Alternative hypothesis: There is difference in the means among the sections of students

Step 2: Significance level
are alpha is set to .05

Step 3: Critical value of F
This is done by the computer and it indicates that the F critical for the treatment (time periods) is 3.49 and the F critical for the blocks (section of students) is 3.26. There are two F criticals because there are two sets of hypotheses, one for the time periods and one for the students.

Step 4: Calculate
The computed F-value for treatment (time periods) is 0.25
The computed F-value for the blocks (section of students) is 0.89

Step 5: Decision
Since the F-value of the treatment (time periods) is 0.25 is less than F critical of 3.49 at an alpha of .05 we do not reject the null hypothesis

Since the F-value of the blocks (section of students) is 0.89 is less than F critical of 3.26 at an alpha of .05 we do not reject the null hypothesis

Step 6: Conclusion
Treatment (Time period)
Since we did not reject the null hypothesis, we can conclude that there is no evidence that time of day affects the quiz scores.

Blocks (Section of Student)
Since we did not reject the null hypothesis, we can conclude that there is no evidence that group affects the quiz scores.

From this, we know that time of day and the group a student belongs to does not matter. If the time of day mattered it might have been due to a host of factors such as early morning or late afternoon. For the groups, the difference could be identified by how they did on individual items. Maybe they struggled with finding the means of question 3.

Remember in this example there was no difference. The ideas above are for determining why there was a difference if that had happened.

One-Way Analysis of Variance (ANOVA)

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Analysis of variance is a statistical technique that is used to determine if there is a difference in two or more sample populations.  Z-test and t-tests are used when comparing one sample population to a known value or two sample populations to each other. When two or more sample populations are involved it is necessary to use analysis of variance.  The simple rule is 3 or more use analysis of variance

Analysis of variance is too complicated to do by hand, even though it is possible. It takes a great deal of time and one error will ruin the answer. Therefore, we are not going to look at equations during this example. Instead, we will focus on the hypotheses and practical applications of analysis of variance. To calculate analysis of variance results you can use SPSS or Microsoft excel.

There are several types of analysis of variance. We are going to first look at one-way analysis of variance.

Here are the assumptions for one-way analysis of variance

  • Samples are randomly selected
  • Samples are independently assigned
  • Samples are homogeneous
  • Sample is normally distributed

One-way analysis of variance is used when 2 or more groups receive the same treatment or intervention. The treatment is the independent variable while the means of each group is the dependent variable. This is because as the researcher, you control the treatment but you do not control the resulting mean that is recorded. One-way analysis of variance is often used in performing experiments.

Let’s look at an example. You want to know if there is any difference in the average life of four different breeds of dogs. You take a random sample of five dogs from four different breeds. Below are the results

Terrier    Retriever   Hound   Bulldog
12                 11                   12            12
13                 10                   11             15
14                 13                   15             10
11                 15                   15             12
15                 14                   16             11

In this example, the independent variable is the breed of dog. This is because you control this. You can select whatever dog breed you want. The dependent variable is the average length of the dog’s lives. You have no control over how long they live. You are trying to see if  dog breed influences how long the dog will live

Here are the hypotheses

Null hypotheses: There is no difference in the average length of a dog’s life because of breed

Alternative hypotheses: There is a difference in the average length of a dog’s life because of breed

The significance level is 0.05  are F critical is 3.24

After running the results in the computer we get an F-value of 0.76. This means we do not reject are null hypotheses.  This means that there is no difference in the average life of the dog breeds in this study.

One-way analysis is used when we have one treatment and three or more groups that experience the treatment. This statistical tool is useful for research designs that call on the need for experiments.

Testing the Difference Between two Means: Paired Samples

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The paired sample t-test is used to compare two sample populations that are correlated. This test is most commonly employed for “before and after” or pretest-postest design. Below are the assumptions of paired sample t-test.

  • Only the matched pairs are used to perform the analysis
  • The data is normally distributed
  • The variances of the two samples are homogeneous
  • The observations are independent of each other

Below are the steps involved in conducting a paired sample t-test

  1. Set up the hypotheses
    • H0: The mean of the paired samples are the same or (μ1 = μ2)
    • H1: The means of the paired samples are not equal or
      (μ1 ≠ μ2, μ1 > μ2, μ1 < μ2)
  2. Determine the level of statistical significance (.1, .05, or .01) and if it is two-tailed or one-tailed
    • Two-tailed means there are two choices. One mean can be greater or lesser than the other.
    • One-tailed means there is one choice. One of the means is greater or it lesser but not vice versa.
  3. You also must take into account the degree of freedom which is sample size – 1 this information is useful when looking at the t-test chart to calculate the t critical value
  4. Calculate the paired t-test. The formulas are below. Take note that there are three separate formulas labeled A, B, and CFORMULA A   t computed = Mean difference
                                                                     standard deviation of the mean /                                                                               square root of the  sample sizeFORMULA B    mean difference = sum of the difference
                                                                                  sample size

FORMULA C    standard deviation of the difference =
√ΣD² – (ΣD)²
                 n     
n – 1
Sorry there is no simple way to explain formula C

4. Make Statistical decision

5. State conclusion

Below is an example

A teacher develops an incentive plan for his students. Students who were quiet got additional stickers in their notebook. The teacher picked 10 students at random to see if the number of stickers they earned was more after the incentive program was adopted. Here are the results

Student               Before             After
1                               20                        35
2                               30                        41
3                                25                       38
4                                31                       42
5                                19                       18
6                                18                       16
7                                23                       34
8                                32                       19
9                                24                       24
10                             19                       33

Step 1 State the hypotheses
H0: μ1 < 0 or the number of stickers after the incentive plan are not more than before
H2: μ1 > 0 or the number of stickers is greater after the incentive plan

Step 2: The level of statistical significance is .05. This is also a one-tailed test

Step 3: Calculate the critical region which is
degrees of freedom = sample size – 1 = tcritical
df = 10 – 1= 9 and the tcritical is 1.83 according to the table

Step 4: Compute t computed for paired samples

Student    Before     After      Difference       Difference²
1                       20                35           15                             225
2                       30                41           11                             121
3                       25                38           13                             169
4                       31                42            11                             121
5                       19                18            -1                                    1
6                       18                16            -2                                    4
7                        23                34           11                              121
8                        32                19         -13                              169
9                        24                24             0                                     0
10                     19                33           14                               196
Sum of difference     59 Sum of the difference² 1127

Find the mean of the difference
59 / 10 = 5.9

Find the standard deviation of the differences (the entire equation below is squared)

1127 – (5.9)²
                10      
10-1

the standard deviation is 11.17

Find the t computed

            59              = 16.69
11.17 / √10

Step 5: Decision

Since the t computed 16.69 is greater than the t critical of 1.83 we reject the null hypothesis

Step 6 Conclusion

Since we reject the null we can conclude that there is evidence that the incentive program has increased the number of stickers the students earn.