Question

compare the three types of t-tests by discussing when each is most appropriate to use and which type of question each type of t-test best to answers. include specific example to illustrate the appropriate use of each test

Answer 1

QUESTION

Compare the three types of t-tests by discussing when each is most appropriate to use and which type of question each type of t-test best to answers. Include specific example to illustrate the appropriate use of each test.

ANSWER

t-TEST

INTRODUCTION

• The t-test is a basic test that is limited to two groups.For multiple groups, you would have to compare each pair of groups, for example with three groups there would be three tests (AB, AC, BC), whilst with seven groups there would need to be 21 tests.
• The basic principle is to test the null hypothesis that the means of the two groups are equal.

The t-test assumes:

1.A normal distribution (parametric data

2. Underlying variances are equal (if not, use Welch's test)

It is used when there is random assignment and only two sets of measurement to compare.

t-TEST

• t –test is about means: distribution and evaluation for group distribution

• Withdrawn from the normal distribution

• The shape of distribution depend on sample size and, the sum of all distributions is a normal distribution

• t- distribution is based on sample size and vary according to the degrees of freedom

SIGNIFICANCE OF t-TEST

• t test is a useful technique for comparing mean values of two sets of numbers.

• The comparison will provide you with a statistic for evaluating whether the difference between two means is statistically significant.

• t test can be used either :

1. To compare two independent groups (independent- samples t test)

2. To compare observations from two measurement occasions for the same group (paired-samples t test).

• The null hypothesis states that any difference between the two means is a result to difference in distribution.

• Remember, both samples drawn randomly form the same population.

• Comparing the chance of having difference is one group due to difference in distribution.

• Assuming that both distributions came from the same population, both distribution has to be equal.

• Then, what we intend:

“To find the he difference due to chance”

• Logically, The larger the difference in means, the more likely to find a significant t test.

• But, recall:

1. Variability

More (less) variability = less overlap = larger difference

2. Sample size

Larger sample size = less variability (pop) = larger difference

TYPES

1. The one-sample t test is used compare a single sample with a population value. For example, a test could be conducted to compare the average salary of nurses within a company with a value that was known to represent the national average for nurses.

2. The independent-sample t test is used to compare two groups scores on the same variable. For example, it could be used to compare the salaries of nurses and physicians to evaluate whether there is a difference in their salaries.

3. The paired-sample t test is used to compare the means of two variables within a single group. For example, it could be used to see if there is a statistically significant difference between starting salaries and current salaries among the general nurses in an organization.

ASSUMPTIONS OF t-TEST

• Dependent variables are interval or ratio.

• The population from which samples are drawn is normally distributed.

• Samples are randomly selected.

• The groups have equal variance (Homogeneity of variance).

• The t-statistic is robust (it is reasonably reliable even if assumptions are not fully met.

ASSUMPTION

1. Should be continuous (I/R)

2. the groups should be randomly drawn from normally distributed and independent populations

e.g. Male X Female

Nurse X Physician

Manager X Staff

NO OVER LAP

3. The independent variable is categorical with two levels

4. Distribution for the two independent variables is normal

5. Equal variance (homogeneity of variance)

6. Large variation = less likely to have sig t test = accepting null hypothesis (fail to reject) = Type II error = a threat to power

APPLICATIONS

1.To compare the mean of a sample with population mean.

2.To compare the mean of one sample with the mean of another independent sample.

3.To compare between the values (readings) of one sample but in 2 occasions.

EXAMPLES

1.Sample mean and population mean

• The general steps of testing hypothesis must be followed.
• Ho: Sample mean=Population mean.
• Degrees of freedom = n – 1

          

−

Example:

The following data represents hemoglobin values in gm/dl for 10 patients:

10.5 9             6.5         8           11

7                7.5 8.5        9.5         12

Is the mean value for patients significantly differ from the mean value of general population.(12 gm/dl) .Evaluate the role of chance.

Solution:

• Mention all steps of testing hypothesis.
•    t=   8.95 − 12          = −5.352

1.80201

  

• Then compare with tabulated value, for 9 df, and 5% level of significance.
• It is = 2.262
• The calculated value>tabulated value.
• Reject Ho and conclude that there is a statistically significant difference between the mean of sample and population mean, and this difference is unlikely due to chance.

2.Two independent samples

• The following data represents weight in Kg for 10 males and 12 females.

Males:      80 75 95 55 60

70 75 72 80 65

Females: 60   70 50 85 45 60

80 65 70 62 77 82

• Is there a statistically significant difference between the mean weight of males and females.Let alpha = 0.01
• To solve it follow the steps and use this equation.

Results:

• Mean 1=72.7       Mean2=67.17
• Variance 1=128.46    Variance2=157.787
• Df = n1+n2-2=20
• t = 1.074
• The tabulated t, 2 sides, for alpha 0.01 is 2.845 .
• Then accept Ho and conclude that there is no significant difference between the 2 means. This difference may be due to chance.
• P>0.01 11

3.One sample in two occasions

• Mention steps of testing hypothesis.
• The df here = n – 1.

Example:

Blood pressure of 8 patients, before & after treatment

BP before               BP after                   d                 square of d(∑d2)

180                               140                       40                      1600

200                              145                        55                     3025

230                              150                        80                     6400

240                              155                       155                     85

170                              120                       50                     2500

190                              130                       60                     3600

200                             140                         60                   3600

165                           130                           35                    1225

Mean d=465/8=58.125

,∑d=465      

∑d2=29175

Results and conclusion:

• t=9.387
• Tabulated t (df7), with level of significance 0.05, two tails, = 2.36
• We reject Ho and conclude that there is significant difference between BP readings before and after treatment.
• P<0.05. 14

CONCLUSION

A t-test is a statistical test that is used to compare the means of two groups. It is often used in hypothesis testing to determine whether a process or treatment actually has an effect on the population of interest, or whether two groups are different from one another.