Hypothesis Testing Lecture 30

 Hypothesis Testing about Mean

Lecture 30

Case II: Hypothesis Testing about μ  when σ is Unknown

Let  X¯ be the unbiased estimate computed from the values of a random sample of size n selected from a normal population having a mean and unknown standard deviation σ. As σ is unknown, so replace by its estimate S. If the sample size is large, the sampling distribution of the sample mean approaches the normal distribution.

Where:

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. Critical Region:

Reject H0 when |Z| ≥ Zα/2

Reject H0, when Z < - Zα 

Reject H0 when Z > Zα

v. Computation:

vi. Remarks

Example 8.4: The average score of a college student is 80. A random sample of 50 students is selected; the sample mean is 87 and the standard deviation is 25. Test the hypothesis that the average score is 80 at the 5% significance level.

Solution:

i. State null & alternative hypothesis

H0: μ = 80 Vs.  H1: μ ≠ 80

ii. The significance level; α = 0.05

iii. The test statistic: If n is large, then

under H0.

iv. Reject H0 when |Z|≥ Z0.5/2 = 1.96

v. Computation:

vi. Remarks: The computed value of the test statistic falls in the rejection region. We do not have sufficient evidence to accept the null hypothesis. Thus, it is concluded that the average score is not 80.

Example8.5: A random sample of 56 observations yielded a sample mean of 70 and a standard deviation of 5.8. At the 10% significance threshold, the population mean is said to be at least 67.

Solution:

i. State null & alternative hypothesis

H0: μ≥ 67  Vs. μ < 67

ii. The significance level; α = 0.10

iii. The test statistic: As n = 56 is large enough, then

vi. Reject H0 when Z < - Z0.10 = - 1.65

v. Computation:

vi. Remarks: The calculated z value, which is 3.87, falls in the critical area. Thus, we do not have sufficient evidence to accept H0 at the 10% significance level. It is concluded that the population mean is less than 67.

Hypothesis Testing about Population Proportion

Let p^ be the estimate of P computed from the values of a random sample of size n selected from a binomial population with proportion. If the sample size is sufficiently large, the sampling distribution of p^ approaches the normal distribution with a mean P that and the standard deviation

.

if it is desired to test H0: P = P0

As x is discrete random variable, using correction factor 

Testing Procedure:

i. State null & alternative hypothesis

H0: P = P0 Vs. H1: P ≠ P0

H0: P ≥ P0 Vs. H1: P < P0

H0: P≤ P0 Vs. H1: P > P0

ii. The significance level; α

iii. The Test Statistic: If n is large enough, then

iv. Critical Region:

Reject H0 when |z| ≥ zα/2

Reject H0 when z < - zα

Reject H0 when z > zα

v. Computation

vi. Remarks

Example 8.6: According to a medical company, the medicine A 75% relieving in flue. The practitioner verified the claim by selecting 120 people at random, and only 38 of them who had flue provided relief by setting a 5% significance level.

Solution: We setup our hypothesis as:

$$ i.  State null & alternative hypothesis

H0: P = 0.75 Vs. H1: P ≠ 0.75

ii. The significance level; α = 0.05

iii. The Test Statistic: If n is large enough, then

iv. Reject H0 when |z| ≥ z0.05/2 = 1.96

v. Computation:

vi. Remarks: The computed z value, which is - 4.53, falls in the rejection area. The sample data do not provide sufficient evidence to accept the hypothesis, and it is concluded that the company claim is not validated. 

Example 8.7: It is observed that 300 out of 700 customers purchased the groceries online. Can we say that least of the customers are moving towards online shopping even for groceries?

Solution: Let P represent the proportion of customers who purchase the groceries online. 

The proportion of customers who purchase and not purchase online groceries is = 1

i. State null & alternative hypothesis

H0: P≥ 0.5 Vs. H1: P < 0.5

ii. The significance level; α = 0.05

iii. The Test Statistic: If n is large enough, then

iv. Reject H0 when z <- z0.05 = - 1.645

v. Computation:

vi. Remarks: The computed z value, which is - 3.74, falls in the rejection area. The sample data do not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that least of the customers purchase online groceries.

Example 8.8: In a study, 172 participants out of 420010 cell phone users went to acquire brain tumors. Examine the hypothesis that the rate of brain cancer among mobile phone users is higher than that of non-users. The rate for the non-cell phone users is less than 0.034%.

Solution: We setup our hypothesis as:

i. State null & alternative hypothesis

H0: P≤ 0.00034 Vs. H1: P > 0.00034

ii. The significance level; α = 0.05

iii. The Test Statistic: If n is large enough, then

iv. Reject H0 when z > z0.05 = 1.645

v. Computation:

vi. Remarks: The computed z value, which is 2.45, falls in the rejection area. The sample data do not provide sufficient evidence to accept the null hypothesis, and it is concluded that the rate of brain cancer among mobile phone users is higher than that of non-users. 

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