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

Introduction to t Distribution Lecture 41

 Introduction 

to 

t-Statistic

Lecture 41

Introduction

The z statistic is used as a test statistic to make inferences about the population mean or the difference between population means (to construct a confidence interval for the population mean, the difference of the population means, or to test the hypothesis about the population means, the difference between the population means) when the population standard deviation is known or unknown and the sample size is large. Now, if making inferences about the population mean or the difference between population means and the population standard deviation(s) is unknown and the sample size is small, the t statistic is used as a test statistic. The t statistic is also known as the student’s t statistic.

The student’s t statistic was developed by William Sealy Gosset in 1908 using the nickname Student. William Sealy also developed the t-test and t-distribution. William Sealy, who worked at the Guinness brewery in Dublin, discovered that the small sample sizes he experienced at work were inappropriate for the statistical methods that were then in use, which used large samples. The student develops the following test statistic to tackle the small sample(s).

Where:

ν: Sample size minus population parameters to be estimated.

n: Sample size.

s: Unbiased estimate of population standard deviation.

Assumptions using Student's t distribution

i. A random selection process will be employed when selecting the sample observations.

ii. The sampled population should be normal. However, a slight departure from normality does not seriously affect the test.

iii. In the case of two samples, both samples are selected randomly and independently from two normal populations having identical variances.

Sampling Distribution of t

The t variable is the ratio of the standard normal variable and the square root of the chi square variable divided by its degree of freedom.

Let X1, X2,..., Xn be the observations of a random sample of size n drawn from a normal distribution with mean μ and standard deviation σ. 

The sampling distribution of t is called t-distribution with (n-1) degree of freedom. The probability function of t is given by

Properties of t-distribution

i. The t-distribution is symmetric and continuous around t = 0 ranges from $-\propto to +\propto .$

ii. The mean of t-distribution is equal to zero. 

μt =E(t)=0.

iii. The variance of the t-distribution is dependent on the degree of freedom and equal to v/v-2.

 σt =Var(t)= v / v-2

iv. The t-distribution is uni-model, and mode of t-distribution is equal to zero.

v. The shape of the t-distribution curve is identical to the normal distribution curve but flatter than the normal curve for a small sample size.

vi. The shape of the t-distribution approach to the normal curve if the sample size is sufficiently large.

vii. The student's t distribution is independent of parameters and depends on the degree of freedom.

vii. The student's t distribution is symmetrical about zero. thus t1-α =- tα

The shape of the student's t distribution and normal distribution curves are given below:

Small Samples Confidence Interval

Confidence Interval for the Mean of a Normal Population When n is small and σ 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 

Now to a 100 (1-α)% confidence interval for μ, choose two values (-tα/2, tα/2) from the table and make the following probability statement.

Example 10.1: The quiz marks of 9 students are given below:

8, 7.5, 6.7, 7, 5.9, 8.2, 6.5, 8.75, 7.65

Find a 95% confidence interval for the students of the whole class. Assume the class population is normal.

Solution:

1-α = 0.95

α = 0.05

tα/2 (v) = t0.025(8) = 2.752

It means 95% confident that the mean score lies between 6.523 and 8.187.

• Read More: Hypothesis Testing on the Test