Hypothesis Testing about Population Mean Lecture 42

 

 Small Samples Test

Lecture 42

Hypothesis Testing about Population Mean when Standard Deviation is unknown

Let X¯ be the unbiased estimate of μ computed from the values of a small random sample of size n selected from a normal population having mean "μ" and unknown standard deviation "σ". As the sample size is small, the sampling distribution of X¯ approaches the t distribution with v = n -1 degree of freedom.

Thats 

Testing Procedure:

i. State null and alternative hypotheses

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. Critical region:

Reject H0 when | t | > tα/2 (v)

Reject H0 when t < - tα(v)

Reject H0 when t > tα (v)

v. Computation:

vi. Remarks.

Example 10.2: According to a private telecom service provider, the average monthly payment for an individual customer is Rs. 500 per month. It is challenging to get in touch with the telecom service provider; therefore, a small random sample of 15 customers is selected, and they are asked about bills they deposit from the telecom service provider. The mean of the 25 customer samples is 510, with a standard deviation of 5. Verify the hypothesis that the telecom service provider’s assertion is accurate.

Solution:

i. State null and alternative hypotheses

H0: μ = 510 Vs. H1: μ ≠ 510

ii. The significance level; α

iii. The test statistic: The small sample test

iv. Critical region:

Reject H0 when | t | > t0.025 (24) = 2.490

v. Computation:

vi. Remarks: The calculated t value falls in the rejection area; the sample information does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that the telecom service provider assertion is not accurate.

Example 10.3: In a particular country, the average height of an adult male is at least 170 cm. It is assumed that some environmental factors may be responsible for the varying average height of the adult male in a particular city in that country. The following height (in centimetres) values are obtained from a random sample of size 9 of the adult males in the city:

176.2 157.9 160.1 180.9 165.1 167.2 162.9 155.7 166.2

Assume that the height distribution is normally distributed. Does H0 have sufficient evidence to be rejected at the significance level α=0.05?

Solution:

i. State null and alternative hypotheses

H0: μ ≥ 170 Vs. H1: μ < 170

ii. The significance level; α = 0.05

iii. The test statistic:

iv. Critical Region

Reject H0 when t < - t 0.05(8) = - 2.306

v. Computation:

vi. Remarks: The calculated t value falls in the acceptance area; the sample information does not provide sufficient evidence to reject the null hypothesis. Thus, it is concluded that the average height of the adult male population is more than 170 centimetres.

Example 10.4: Suppose that Scholastic Aptitude Test scores have a normal distribution with m = 500. To increase SAT scores, a high school counsellor created a unique course. 16 students are chosen at random to enrol in the course and subsequently take the SAT. The sample had an average score of 544 and a standard deviation of 45. Does taking the course improve one’s SAT scores? Test at a = 0.05.

Solution:

i. State null and alternative hypotheses

H0: μ ≤ 500 Vs. H1: μ > 500

ii. The significance level; α = 0.05

iii. The test statistic:

iv. Critical Region

Reject H0 when t > t0.05(15) = 2.131

v. Computation:

vi. Remarks: The calculated t value falls in the rejection area; the sample information does not provide sufficient evidence to accept the null hypothesis. Therefore, it can be said that the course raised the SAT score.

Example 10.5: A corporation firm desires to verify the claim that their batteries have a life span of over 40 hours. To test the claim, a random sample of 15 batteries produced a mean of 44.9 and an 8.9 standard deviation. Use a level of significance of 0.05 to test the corporation's claim.

Solution:

i. State null and alternative hypotheses

H0: μ ≤ 40 Vs. H1: μ > 40

ii. The significance level; α = 0.05

iii. The test statistic:

iv. Critical Region

Reject H0 when t > 0.05(14) = 2.145

v. Computation:

vi. Remarks: The calculated t value falls in the acceptance area; the sample information does not provide sufficient evidence to reject the null hypothesis. Thus, it is concluded that the average life span is less than 40 hours. 

• Read More: Two Samples t Test