Significance (Hypothesis) Testing

Hypothesis testing is the process of making statements about a population parameter on the basis of sample evidence. Here, we test whether a set of sample results are consistent with some fact or supposition about the population. For example, we may have a survey which is carried out annualy and we wish to test if figures have changed since last year.

The idea tested is called the null hypothesis (H0), and it could be either rejected or accepted. The alternative hypothesis to that of null hypothesis is labelled (H1).

The null hypothesis can be tested using confidence interval: if H0 was contained within the confidence interval it would be accepted, otherwise it is rejected.

Thus, the 95% confidence interval will be: μ = Â± 1.96 s/ √n.

Thus, the 99% confidence interval will be: μ = Â± 2.576 s/ √n.

For a population mean H0 : Âµ = Âµ0 ; H1 : Âµ ≠ Âµ0 ;

For a population percentage H0 : π = π 0 ; H1 : π ≠ π0 ;

Throughout the test H0 is assumed to be true, and the basic layout and procedure will be:

Step:1. State hypothesis2. State significance level3. State critical (cut"“off) values4. Calculate the test statistic (Z)5. Compare the Z value to the critical value6. Come to conclusion7. Conclude using English Example:H0 : Âµ = Âµ0 ; H1 : Âµ ≠ Âµ0 ; 5%; 1%"“1.96, +1.96; "“2.576, +2.576varies for each testreject or accept H0

For a test statistic for a population mean, we use the formula:

Example:

It is believed that the average amount spent per week on jam is Â£1.50. surveying a random sample of 80 shoppers selected from a large population have shown that an average of Â£1.40 is spent weekly on jam, with a standard deviation of Â£0.15.

Is the original belief supported by these results at the 5% significance level?

Solution:

H0 : Âµ = Âµ0 =1.50;

H1 : Âµ ≠ Âµ0 ≠1.50;

Significance level is 5%;

The critical values are: "“1.96 and +1.96;

The calculated value ("“5.96) is below the lower critical value ("“1.96). Therefore, the null hypothesis is rejected at the 5% level of significance. In other words, the sample evidence does not support the original belief the shoppers spend an average of Â£1.50 per week on jam. The amount spent is significantly different from Â£1.50.

To test a population percentage, we use the formula: .

The sampling error is :

Example:

An auditor claims that 10% of invoices for a company are incorrect. To test this claim a random sample of 100 invoices are checked, and 12 are found to be incorrect.

Test at the 5% significance level, if the auditor's claim is supported by the sample evidence.

Solution:

H0 : π = π0 = 10%;

H1 : π ≠ π0 ≠10%;

Significance level is 5%;

The critical values are: "“1.96 and +1.96;

The sample percentage is (12/ 100)* 100 = 12%

The calculated value falls between the two critical values, so we cannot reject the null hypothesis. Thus, the evidence from the sample is consistent with the auditor's claim that 10% of the invoices are incorrect.

One"“Sided Significance Test:

In these tests we are able to specify whether the real value is above or below the claimed value in cases of rejected null hypothesis.

To test a decrease in a survey figures:

H0 : π = π 0 ; H1 : π < π0 ;

To test a decrease in a survey figures:

H0 : π = π 0 ; H1 : π > π0 ;

In this test the chance of rejecting H0 Will be concentrated at one end of the Normal distribution. Where the significance level is 5%, the critical value will be:

Â· "“1.645 for H0 : π = π0 ; H1 : π < π0 ;

Â· +1.645 for H0 : π = π0 ; H1 : π > π0 ;

If the significance level is 1%, the critical value becomes either "“2.33 or + 2.33.

A one"“sided test for a population mean will be:

H0 : Âµ = Âµ0 ; H1 : Âµ > Âµ0 ;

H0 : Âµ = Âµ0 ; H1 : Âµ < Âµ0 ;

But for a population percentage , the one"“sided test will be:

Example:

A company claims that 4% of the components supplied by a certain supplier are defective,. The supplier disputes this claim. To test the validity of this claim, a random sample of 500 components was selected, and each one was carefully checked. The results showed that 12 were defective.

Use an appropriate test, at the 5% level, to find out if the company's claim is valid.

Solution:

H0 : π = π0 = 4%;

H1 : π < π0 <4%;

Significance level is 5%;

The critical values are: "“1.645;

The sample percentage is (12/ 500)* 100 = 2.4%

"“1.862 < "“1.645 so H0 is rejected, meaning that the sample evidence suggests that the company is wrong in claiming that 4% of the suppliers' components are defective.

However, the one"“sided test can be used to test the producer's risks by using:

H1 : Âµ > Âµ0 . Also, it can be used to test consumers' risks by using: H1 : Âµ < Âµ0 .

Types of Error:

There will always be some chance that the true population value does really lie outside of the confidence interval.

Accept H0 Reject H0

If H0 is correct Correct decision Type I error

If H0 is incorrect Type II error Correct decision

Type I Error: is the rejection of H0 when it is true. This probability is the significance level of the test (5% or 1%), and it is decided before the test is conducted. However, this error can be controlled.

Type II Error: is the failure to reject H0 when it is false. This probability cannot be determined before the test is conducted.

Hypothesis Testing with two Samples:

1. Testing For A Difference Of Means:

Here, we assume that the two samples come from the same population, and therefore, the two population means will be the same as follows:

H0: Âµ1 = Âµ2 or : Âµ1 "“ Âµ2 = 0

If it is a two"“sided test: H1: Âµ1 ≠Âµ2 or : Âµ1 "“ Âµ2 ≠ 0

If it is a one"“sided test: H1: Âµ1 >Âµ2 or : Âµ1 "“ Âµ2 > 0 or:

H1: Âµ1 <Âµ2 or : Âµ1 "“ Âµ2 < 0

Accordingly, the test statistic will be:

If H0 says that the two population means are equal, the following formula is to be used:

2. Testing For A Difference In Population Percentage:

The test statistic will be:

If H0 says that the two population percentages are equal, the following formula is used:

Hypothesis Testing with Small Samples:

The Normal distribution only holds for large samples; therefore for small samples we use the t"“distribution.

For a single sample, the test statistic for a population mean is: with (n"“1) degrees of freedom. But for a population percentage, we use:

with (n"“1) degrees of freedom.

Example:

A company claims that the average annual maintenance for its vehicles is Â£500. There is some belief that the average is higher then what is claimed to be. To test this, a sample of 6 cars was randomly selected. The mean annual maintenance cost was found to be Â£555, with a standard deviation of Â£75.

Use an appropriate test, at the 5% level, to find if the company's claim is valid.

Solution:

H0 : Âµ = Â£500;

H1 : Âµ >Â£500;

Significance level is 5%;

Degrees of freedom = 6 "“ 1= 5

The critical value is: 2.015.

1.796 < 2.015; therefore, H0 cannot be rejected, meaning that the sample evidence does not suggest that maintenance costs are more than Â£500 per annum.

For two samples, the pooled standard error will be:

with (n1 + n2 "“ 2 ) degrees of freedom.

Example:

A company has two factories, one in the UK and one in Germany. The head office staff feel that the German factory is more efficient than the British one, and to test this two random samples were selected. The British sample has 20 workers who take an average of 25 minutes to complete a standard task. Their standard deviation is 5 minutes. The German one has 10 workers who take an average of 20 minutes toi complete the same task, with a standard deviation of 4 minutes.

Use an appropriate hypothesis test, at the 1% level, to find if the German workers are more efficient.

Solution:

H0 : (Âµ1 "“ Âµ2) = 0;

H1 : (Âµ1 "“ Âµ2) > 0;

Significance level is 1%;

Degrees of freedom = 20 + 10 "“ 2= 28

The critical value is: 2.467.

2.746 > 2.467, therefore, the null hypothesis is rejected, and it appears that the German workers are more efficient at this particular task.

Hypothesis testing is the process of making statements about a population parameter on the basis of sample evidence. Here, we test whether a set of sample results are consistent with some fact or supposition about the population. For example, we may have a survey which is carried out annualy and we wish to test if figures have changed since last year.

The idea tested is called the null hypothesis (H0), and it could be either rejected or accepted. The alternative hypothesis to that of null hypothesis is labelled (H1).

The null hypothesis can be tested using confidence interval: if H0 was contained within the confidence interval it would be accepted, otherwise it is rejected.

Thus, the 95% confidence interval will be: μ = Â± 1.96 s/ √n.

Thus, the 99% confidence interval will be: μ = Â± 2.576 s/ √n.

For a population mean H0 : Âµ = Âµ0 ; H1 : Âµ ≠ Âµ0 ;

For a population percentage H0 : π = π 0 ; H1 : π ≠ π0 ;

Throughout the test H0 is assumed to be true, and the basic layout and procedure will be:

Step:1. State hypothesis2. State significance level3. State critical (cut"“off) values4. Calculate the test statistic (Z)5. Compare the Z value to the critical value6. Come to conclusion7. Conclude using English Example:H0 : Âµ = Âµ0 ; H1 : Âµ ≠ Âµ0 ; 5%; 1%"“1.96, +1.96; "“2.576, +2.576varies for each testreject or accept H0

For a test statistic for a population mean, we use the formula:

Example:

It is believed that the average amount spent per week on jam is Â£1.50. surveying a random sample of 80 shoppers selected from a large population have shown that an average of Â£1.40 is spent weekly on jam, with a standard deviation of Â£0.15.

Is the original belief supported by these results at the 5% significance level?

Solution:

H0 : Âµ = Âµ0 =1.50;

H1 : Âµ ≠ Âµ0 ≠1.50;

Significance level is 5%;

The critical values are: "“1.96 and +1.96;

The calculated value ("“5.96) is below the lower critical value ("“1.96). Therefore, the null hypothesis is rejected at the 5% level of significance. In other words, the sample evidence does not support the original belief the shoppers spend an average of Â£1.50 per week on jam. The amount spent is significantly different from Â£1.50.

To test a population percentage, we use the formula: .

The sampling error is :

Example:

An auditor claims that 10% of invoices for a company are incorrect. To test this claim a random sample of 100 invoices are checked, and 12 are found to be incorrect.

Test at the 5% significance level, if the auditor's claim is supported by the sample evidence.

Solution:

H0 : π = π0 = 10%;

H1 : π ≠ π0 ≠10%;

Significance level is 5%;

The critical values are: "“1.96 and +1.96;

The sample percentage is (12/ 100)* 100 = 12%

The calculated value falls between the two critical values, so we cannot reject the null hypothesis. Thus, the evidence from the sample is consistent with the auditor's claim that 10% of the invoices are incorrect.

One"“Sided Significance Test:

In these tests we are able to specify whether the real value is above or below the claimed value in cases of rejected null hypothesis.

To test a decrease in a survey figures:

H0 : π = π 0 ; H1 : π < π0 ;

To test a decrease in a survey figures:

H0 : π = π 0 ; H1 : π > π0 ;

In this test the chance of rejecting H0 Will be concentrated at one end of the Normal distribution. Where the significance level is 5%, the critical value will be:

Â· "“1.645 for H0 : π = π0 ; H1 : π < π0 ;

Â· +1.645 for H0 : π = π0 ; H1 : π > π0 ;

If the significance level is 1%, the critical value becomes either "“2.33 or + 2.33.

A one"“sided test for a population mean will be:

H0 : Âµ = Âµ0 ; H1 : Âµ > Âµ0 ;

H0 : Âµ = Âµ0 ; H1 : Âµ < Âµ0 ;

But for a population percentage , the one"“sided test will be:

Example:

A company claims that 4% of the components supplied by a certain supplier are defective,. The supplier disputes this claim. To test the validity of this claim, a random sample of 500 components was selected, and each one was carefully checked. The results showed that 12 were defective.

Use an appropriate test, at the 5% level, to find out if the company's claim is valid.

Solution:

H0 : π = π0 = 4%;

H1 : π < π0 <4%;

Significance level is 5%;

The critical values are: "“1.645;

The sample percentage is (12/ 500)* 100 = 2.4%

"“1.862 < "“1.645 so H0 is rejected, meaning that the sample evidence suggests that the company is wrong in claiming that 4% of the suppliers' components are defective.

However, the one"“sided test can be used to test the producer's risks by using:

H1 : Âµ > Âµ0 . Also, it can be used to test consumers' risks by using: H1 : Âµ < Âµ0 .

Types of Error:

There will always be some chance that the true population value does really lie outside of the confidence interval.

Accept H0 Reject H0

If H0 is correct Correct decision Type I error

If H0 is incorrect Type II error Correct decision

Type I Error: is the rejection of H0 when it is true. This probability is the significance level of the test (5% or 1%), and it is decided before the test is conducted. However, this error can be controlled.

Type II Error: is the failure to reject H0 when it is false. This probability cannot be determined before the test is conducted.

Hypothesis Testing with two Samples:

1. Testing For A Difference Of Means:

Here, we assume that the two samples come from the same population, and therefore, the two population means will be the same as follows:

H0: Âµ1 = Âµ2 or : Âµ1 "“ Âµ2 = 0

If it is a two"“sided test: H1: Âµ1 ≠Âµ2 or : Âµ1 "“ Âµ2 ≠ 0

If it is a one"“sided test: H1: Âµ1 >Âµ2 or : Âµ1 "“ Âµ2 > 0 or:

H1: Âµ1 <Âµ2 or : Âµ1 "“ Âµ2 < 0

Accordingly, the test statistic will be:

If H0 says that the two population means are equal, the following formula is to be used:

2. Testing For A Difference In Population Percentage:

The test statistic will be:

If H0 says that the two population percentages are equal, the following formula is used:

Hypothesis Testing with Small Samples:

The Normal distribution only holds for large samples; therefore for small samples we use the t"“distribution.

For a single sample, the test statistic for a population mean is: with (n"“1) degrees of freedom. But for a population percentage, we use:

with (n"“1) degrees of freedom.

Example:

A company claims that the average annual maintenance for its vehicles is Â£500. There is some belief that the average is higher then what is claimed to be. To test this, a sample of 6 cars was randomly selected. The mean annual maintenance cost was found to be Â£555, with a standard deviation of Â£75.

Use an appropriate test, at the 5% level, to find if the company's claim is valid.

Solution:

H0 : Âµ = Â£500;

H1 : Âµ >Â£500;

Significance level is 5%;

Degrees of freedom = 6 "“ 1= 5

The critical value is: 2.015.

1.796 < 2.015; therefore, H0 cannot be rejected, meaning that the sample evidence does not suggest that maintenance costs are more than Â£500 per annum.

For two samples, the pooled standard error will be:

with (n1 + n2 "“ 2 ) degrees of freedom.

Example:

A company has two factories, one in the UK and one in Germany. The head office staff feel that the German factory is more efficient than the British one, and to test this two random samples were selected. The British sample has 20 workers who take an average of 25 minutes to complete a standard task. Their standard deviation is 5 minutes. The German one has 10 workers who take an average of 20 minutes toi complete the same task, with a standard deviation of 4 minutes.

Use an appropriate hypothesis test, at the 1% level, to find if the German workers are more efficient.

Solution:

H0 : (Âµ1 "“ Âµ2) = 0;

H1 : (Âµ1 "“ Âµ2) > 0;

Significance level is 1%;

Degrees of freedom = 20 + 10 "“ 2= 28

The critical value is: 2.467.

2.746 > 2.467, therefore, the null hypothesis is rejected, and it appears that the German workers are more efficient at this particular task.