Tests of Hypothesis

A Presentation in Statistics

By: Rufo Tuddao

St. Paul University Philippines

HYPOTHESISA statement or tentative theory which

aims to explain facts about the real world.

An educated guess.It is subject to testing. If it is found to

be statistically true, it is accepted, if not rejected.

Kinds of Hypothesis Null hypothesis (Ho): it serves as the

working hypothesis. It is that which one hopes to accept or reject. It must always express the idea of nonsignificance of difference.

Alternative hypothesis (Ha): it generally represents the hypothetical statement that the researcher wants to prove.

Type I and Type II errors

When making a decision about a proposed hypothesis based on the sample data, one runs the risk of making an error. The following table below summarizes the possibilities:

Decision Actual ConditionHo is true

Actual ConditionHa is true

Reject Ho Type I error Correct decision

Accept Ho Correct decision

Type II error

Level of Significance The probability of making a type I

error or alpha error in a test is called the significance level of the test. The significance level of a test is the maximum value of the probability of rejecting the null hypothesis Ho when in fact it is true.

Example:

A level of significance of .05 signifies that one is 95 % confident that he has made the right decision and allocates only an error of 5 %.

Steps in Hypothesis Testing

Step 1: Formulate the null hypothesis (Ho) that there is no significant difference between items being compared. State the alternative hypothesis (Ha) which is used in case Ho is rejected.

Steps in Hypothesis Testing

Step 2: Set the level of significance of the test,

Step 3: Determine the test to be used. Use the z-test if population standard deviation is given, and t-test if the standard deviation given is from the sample.

Steps in Hypothesis Testing

Step 4: Determine the tabular value for the test. For a z-test, use the table of critical values of z based on the area of the normal curve

Critical Value of z at Varying Significance Levels

Type of Test

Level of Significance

.10 .05 .025 .01

One-tailed test±

1.28±

1.645±

1.96±

2.33

Two-tailed test±

1.645±

1.96±

2.33±

2.58

Steps in Hypothesis Testing

For a t-test, one must first compute for the degrees of freedom (df) then look for the tabular value from the table t-distribution. For a single sample df = n – 1 and for two samples df = n1 + n2

– 2

Steps in Hypothesis Testing

Step 5: Compute for z or t as needed, using any of the following formulas

Z - TEST

It is used if the standard deviation of the population is known or given.

Sample mean (X) compared with population mean ()

Formula #1 (X - ) n where X = sample meanZ = ------------ = population mean = population standard deviation n = number of samples

Comparing two sample means (X1 and X2)

(X1 -X2)Z = ---------------------------

(1/n1) + (1/n2) where X1 = mean of the first sample X2 = mean of the second sample

= population standard deviation n1 = number of items in the first sample n2 = number of items in the second sampleFormula #2

Comparing two sample proportions (p1 and p2)

p1 - p2 where:Z = ---------------- p1 = proportion of the first sample p1q1 + p2q2 p2 = proportion of the second sample ----- ----- n1 = number of items in the 1st sample n1 n2 n2 = number of items in the 2nd sample q1 = 1 – p1

q2 = 1 – p2

Formula #3

T - TEST

It is used if the standard deviation of the sample is known or given.

Sample mean (X) compared with population mean ()

Formula #1 (X - ) where X = sample meant = ------------ = population mean s / n-1 s = sample standard deviation n = number of samples

X1 – X2

t = ------------------------------------------------------------------- (n1 – 1) (s1)

2 + (n2 – 1 ) (s2)2 1 1

------------------------------------ ------ + ------ n1 + n2 – 2 n1 n2

where X1 = mean of the first sample X2 = mean of the second sample s1 = standard deviation of the first sample

s2 = standard deviation of the second sample

n1 = number of items in the first sample n2 = number of items in the second sampleFormula #2

Comparing two sample means (X1 and X2)

Steps in Hypothesis Testing Step 6: Compare the computed value with its

corresponding tabular value, then state your conclusion based on the following guidelines:

  A: Reject Ho if the absolute computed value is

equal to or greater than the absolute tabular value.   B: Accept Ho if the absolute computed value is

less than the absolute tabular value.

Example 1

Data from a school census show that the mean weight of college students was 45 kilos with a standard deviation of 3 kilos. A sample of 100 college students were found to have a mean weight of 47 kilos. Are the 100 college students really heavier than the rest using 0.05 level of significance?

Steps:1. Ho:The 100 college students are not really heavier

than the rest (X= ). Ha: The 100 college students are really heavier

than the rest (X> ).2. Set = .053. The standard deviation of the population is given

therefore the z-test formula #1 is to be used.4. The tabular value for a z-test at .05 level of significance

in a one – tailed test is ±1.645.5.   The given values in the problem are:

X = 47 kilos = 45 kilos = 3 kilos n = 100 students

Example 1Formula # 1 is to be used: 

(X - ) n (47 – 45) 100 2 (10)Z = ------------ = ------------------ = ----------- 3 3

  = 6.67The computed value of 6.67 is greater than the tabular value

which is 1.645, therefore the null hypothesis is rejected.Decision: The 100 college sampled students are really heavier

than the rest of the students.

Example 2 A researcher wishes to find out whether or not there is

significant difference between the monthly allowance of morning and afternoon students in his school. By random sampling, he took a sample of 239 students in the morning session. These students were found to have a mean monthly allowance of P142.00. The researcher also took a sample of 209 students in the afternoon session. They were found to have a mean monthly allowance of P148.00. The total population of students in that school has a standard deviation of P40.00. Is there a significant difference between the two samples at .01 level of significance?

Example 21. Ho: There is no significant difference between the

samples. (X1=X2) Ha: There is a significant difference between the

samples. (X1X2)2. Set = .013. The standard deviation of the population is given

therefore the z-test formula # 2 is to be used.4. The tabular value for a z-test at .01 level of significance

is in a two- tailed test is ±2.58.The given values in the problem are:

X1 = 142 x2 = 148

= 40 n1 = 239 n2 = 209

Example 2Formula # 2 is to be used:

The computed value of | -1.583 | is less than the tabular value which is 2.58, therefore the null hypothesis is accepted.

Decision: There is no significant difference between the two samples.

Question: When do you consider a test is one – tailed or two –

tailed test?Answer: The formulation of the

alternative hypothesis (ha) determines whether the test is one-tailed or two-tailed test. If it is a directional test, then it is a one-tailed test. If it is a non – directional test then it is a two – tailed test.

Examples for a one - tailed test

• 1. Ha: Group A is really brighter than the group B. A>B.

• 2. Ha:Method A is more effective than Method B. A>B

• 3. Ha: The mean marrying age of adults in locality A is really lower than the mean marrying age of adults in locality B. XA< XB

Guiding principle for a one –tailed test

• If there is an adjective used in (Ha) such as taller than, heavier than, more effective, lower than, brighter than etc. than it is a one tailed test.

Examples for a two - tailed test

• 1. Ha: Group A is significantly different from group B. A B.

• 2. Ha:Method A is significantly different from Method B. A B

• 3. Ha: The mean marrying age of adults in locality A is significantly different from the mean marrying age of adults in locality B. XA XB

Guiding principle for a two –tailed test

• If there is a phrase significantly different from and no adjective used in (Ha) such as taller than, heavier than, more effective, lower than, brighter than etc. than it is a two - tailed test.

Example 3A sample survey of a television

program in Metro Manila shows that 80 out of 200 men and 75 out of 250 women dislike the “Rosalinda” program. One likes to know whether the difference between the two sample proportions, 80/200 = .40 and 75/250 = .30, is significant or not at .05 level of significance.

Example 31. Ho: There is no significant difference between the two

sample proportions. (P1 = P2) Ha: There is a significant difference between the two

sample proportions. (P1 P2)2. Set = .053. Use the z-test formula #3 to compare two sample

proportions. 4 The tabular value for a z-test at .05 level of significance

in a two - tailed testis ±1.96.5. The given values in the problem are:  where:  p1 = .40 p2 = .30 n1 = 200, n2 = 250 q1 = 1-.40 = .60 q2 = 1 - .30

=.70

Example 3

The computed value of 2.22 is greater than the tabular value which is 1.96, therefore the null hypothesis is rejected.

Decision: There is a significant difference between men and women viewership.

Example 4A researcher knows that the average

height of Filipino women is 1.525 meters. A random sample of 26 women was taken and was found to have a mean height of 1.56 meters, with a standard deviation of .10 meters. Is there reason to believe that the 26 women in the sample are significantly taller than the others at .05 significance level?

Example 41. Ho: The 26 women in the sample are not significantly taller than

the others (X1= ) Ha: The 26 women in the sample are significantly taller than the

others . (X1 )2. Set = .053. The standard deviation of the sample is given therefore the t-test is

to be used formula # 1.4. Df = n-1 = 26-1 = 25. Therefore, the tabular value of t-test at .05

level of significance with df = 25 in a one - tailed test is. ±1.708.The given values in the problem are:X = 1.56 = 1.525

s = .10 n1 = 26

Example 4

Compute the value of t using formula #1.If the computed value of t is greater than or equal to the absolute tabular value which is ±1.708

reject the null hypothesis. However, if the computed value of t is less than the absolute tabular value which is ±1.708 , accept the null hypothesis.

Example 5 A teacher wishes to test whether or not the Case

Method of teaching is more effective than the Traditional Method. She picks two classes of approximately equal intelligence (verified through an administered IQ test). She gathers a sample of 18 students to whom she uses the Case Method and another sample of 14 students to whom she uses the Traditional Method. After the experiment, an objective test revealed that the first sample got a mean score of 28.6 with a standard deviation of 5.9, while the second group got a mean score of 21.7 with a standard deviation of 4.6. Based on the result of the administered test, can we say that the Case Method is more effective than the Traditional Method at .01 significance level?

Example 51. Ho: The Case Method is as effective as theTraditional Method .

(CM=TM) Ha: The Case Method is more effective than theTraditional Method . (X1X2)2. Set = .013. The standard deviation of the sample is given therefore the t-test

formula # 2 is to be used.4. Df = n1 + n2 -2 = 18 + 14 -2 = 30. Therefore, the tabular value of t-

test at .05 level of significance with df = 30 in a one - tailed test is. ±2.457.The given values in the problem are:X1 = 28.6 x2 = 21.7 s1 =

5.9 s2 = 4.6 n1 = 18 n2 = 14

Example 5

Compute the value of t using formula #1.If the computed value of t is greater than or equal to the absolute tabular value which is ±2.457

reject the null hypothesis. However, if the computed value of t is less than the absolute tabular value which is ±1.708 , accept the null hypothesis.

Assignmment # 2

Direction: Solve the following problems by following the sequential

steps of hypothesis testing. Submit this on August 10, 2008. Good luck

friend.

Problem #1

• It is known from the records of the city schools that the population standard deviation of mathematics test scores on the XYZ test is 5. A sample of 200 pupils from the system was taken and it was found out that the sample mean score is 75. Previous tests showed the population mean to be 70. Is it safe to conclude that the sample is significantly different from the population at .01 level?

Problem #2

• A company is trying to decide which of two types of tires to buy for their trucks. They would like to adopt brand C unless there is some evidence that Brand D is better. An experiment was conducted where 16 tires from each brand were used. The tires were run under similar conditions until they wore out. The results are:

• Brand C: X1 = 40,000 kms., s1 = 5,400 kms

• Brand D: X2 = 38,000 kms., s2 = 3,200 kms.

What conclusions can be drawn at .05 level?

Problem #3

• A sample survey on the average total yearly expenditure included 150 students of a certain university. The mean total expenditure per student per year for the sample was P3,000 with a standard deviation of P500. How likely is it that the students spend an average of P3,500 per year as claimed by a parent at .01 significance level.

Problem #4

• Incoming freshmen are given entrance examinations in a number of fields, including English. Over a period of years, it has been found that the average score in the English examination is 80 with a standard deviation of 7.8. An English instructor examines the scores for his class of a sample of 30 students and finds that their average is 85. Can the instructor claim that the average score has increased at .01 level of significance?

Problem #5

• Two methods of teaching Statistics are being tried by a professor. A class of 49 students is taught by Method A and a class of 36 is taught by Method B. The two classes are given the same final examination. The scores are: X1 = 77, X2 = 80. Using a .01 significance level, can we conclude that the average final examination scores produced by the two methods are different if the population standard deviation is 5?

Problem #6

• A manufacturer packs sugar into plastic bags. Each bag is to hold 5 kilos of sugar. When the production process is under control, each bag contains on the average 5 kilos. At one period, a sample of 17 bags was taken to check the process and was found to weigh 5.2 kilos with a standard deviation of .75 kilos. Is the manufacturing process under control at .05 level?

Problem #7

• All freshmen in a particular school were found to have a variability in grades expressed as a standard deviation of 3. Two samples among these freshmen, made up of 20 and 50 students each, were found to have means of 88 and 85, respectively. Based on their grades, is the first group really brighter than the second group at = .01?