Hypothesis Testing Lecture 28

 Hypothesis Testing

Lecture 28

Hypothesis

The word hypothesis, derived from the ancient Greek word, which means “to put under” or “to suppose,” is the Greek word from which the word hypothesis (plural hypotheses) is derived. A hypothesis is a tentative statement put forward to explain a phenomenon or a real-world problem.

A hypothesis may be defined as a proposed explanation for a phenomenon or logically conjectured relationship between two or more variables, expressed in the form of a testable statement.

Example I: The label on a can of powder milk states that it contains 10 ounces.

Example II: The average height of college students is 62 inches.

Types of Hypotheses

1. Null Hypothesis

2. Alternative Hypothesis

Null Hypothesis

Any hypothesis that is to be tested for possible rejection under the assumption that it is true. The null hypothesis is denoted by H0 and contains some sort of equalities, i.e.,=,≥, ≤

Alternative Hypothesis

The alternative hypothesis that is contradictory to the null hypothesis and denoted by H1 and contains Strick inequalities, i.e.,≠, >, <

Example:

The label on a can of powder milk states that it contains 10 ounces.

Significance Level

Traditionally, 𝜶 is the probability that is used as a standard to reject a null hypothesis when 𝐻0 is taken to be true. The values of 𝜶 that are most frequently used are 0.05, 0.10, and 0.01.

Test Statistic

A statistic that offers evidence against a null hypothesis and a foundation for testing it. The common test statistics are z statistic, t statistic, and F statistic.

Critical Region

It is the part of the sample space (critical region) where the null hypothesis H0 is rejected. 

One-Tailed Test

A test for which the entire critical region is located on only one side of the two tails of the sampling distribution of the test statistic. A one-tailed test is used when the alternative hypothesis is formulated as H1: μ>μ0 or H1: μ<μ0.

Two-Tailed Test

A test for which the entire critical region is located in both sides of the two tails of the sampling distribution of the test statistic. A two-tailed test is used when the alternative hypothesis is of the form H1:μ≠μ0.

TYPE 1 ERROR

When a true null hypothesis is rejected, then type error is committed. The type I is also called false positive error. The probability of type I error is denoted by α. It usually equals the significance level of a test.

$$

α =P[RejectH0 /H0 is true]

 α = P[X¯> c/H0 is true]

TYPE II ERROR

When a false null hypothesis is accepted, while the alternative hypothesis is true. The probability of a type II error is denoted by β.

β =P[Accept H0/H1 is true]

 β = P[X¯> c/H1 is true]

The Power of a Test

The power of a test with respect to a specified alternative hypothesis is the probability of rejecting a true null hypothesis when it is actually false.The power test is the complement of β.

π = P[Reject H0 / H0 is false}

we know that 

β= P [Reject H0 / H0 is false}

 P[Reject H0 / H0 is false}+ P[Reject H0 / H0 is false}=1

β + π = 1

π=1 -β

General Procedure for Hypothesis Testing

The procedure for testing hypotheses about population parameters involves the following six steps:

i. state your problem and formulate an appropriate null with an alternative hypothesis.

ii. Decide upon a significance level, 

iii. Choose an appropriate test statistic and sketch the sampling distribution of the test statistic

iv. Determine the critical region

v. Compute the value of the test statistic from the sample data

vi. Conclusions & Remarks.

TESTS BASED ON NORMAL DISTRIBUTION

Case I: Hypothesis Testing about  when σ is known

Let X¯ be the unbiased estimate of μ computed from the values of a random sample of size n selected from a normal population having a mean and known standard deviation.σ  is known, the sampling distribution of sample mean approaches normal distribution with mean “μ” and standard deviation, whether the sample size is large or small.

If it is desired to test H0: μ = μ0

Testing Procedure:

i. State null & alternative hypothesis

H0: μ = μ0 Vs. H1: μ ≠ μ0

H0: μ ≤ μ0 Vs. H1: μ > μ0

H0: μ ≥ μ0 Vs. H1: μ < μ0

ii. The significance level; α

iii. The test statistic:

iv. The critical region:

Reject H0 if |z|≥ zα/2 (when H1: μ ≠ μ0)

Reject H0 if z> zα (when H1: μ > μ0)

Reject H0 if z< -zα (when H1: μ < μ0)

v. Computation:

vi. Conclusions / Remarks

Example8.1: It is stated that the standard deviation is 6 and the average weight in a certain region is 75 kg. To investigate the claim, an investigator chose a random sample of 16 people, with a sample mean of 82. Which conclusion should be researched?

 

Solution: The population mean will be tested here. 

$$i. State null & alternative hypothesis

H0: μ = 75 Vs. H1: μ ≠ 75

ii. The significance level; α = 0.05

iii. The test statistic: The population standard deviation is known, then the z statistic is used as the test statistic. The test statistic under H0 is given by:

iv. The critical region:

Reject H0 if |z|≥ z0.05/2 = 1.96

v. Computation:

vi. Conclusion: The calculated z value is inside the rejection area. There is insufficient evidence for us to believe H0. Consequentially, it is determined that the assertion about the average weight of the population is not true.

Example 8.2: A telecom service provider claims that individual customers pay on average at least Rs. 400 per month with a standard deviation of Rs. 25. A random sample of 50 customers’ payments during a given month is taken with a mean of Rs. 250. What to say with respect to the claim made by the service provider?

Solution:

$$i. State null & alternative hypothesis

H0: μ ≥ 400 Vs. H1: μ < 400

ii. The significance level; α = 0.05

iii. The test statistic: The population standard deviation is known, then the z statistic is used as the test statistic. The test statistic under H0 is given by:

iv. The critical region:

Reject H0 if z< -z0.05 = -1.645

v. Computation:

vi. Conclusion: The computed z value falls in the rejection region. We have not sufficient evidence to accept H0. Thus. It is concluded that the claim of the telecom company about average payment is not validated.

Example 8.3: An automobile tire manufacturer claims that the average life of a particular grade of tire is more than 20,000 with a standard deviation of 5000 km. A random sample of 16 tires has a mean 22,000 km.

Solution:

$$i. State null & alternative hypothesis

H0: μ ≤ 2000 Vs. H1: μ > 2000

ii. The significance level; α = 0.05

iii. The test statistic: The population standard deviation is known, then the z statistic is used as the test statistic. The test statistic under H0 is given by:

iv. Reject H0 if z> z0.05 = 1.645

v. Computation:

vi. Remarks: The computed z value falls in the accepts region. We have not sufficient evidence to reject Thus, it is concluded that the average life of tiers is less or equal than 20, 000.

• Read More: Hypothesis Testing about Population Proportion