Question

Dr. McDonald thinks that his group of earth science students is particularly special (in a good way), and he is interested in knowing if their class average falls within the boundaries of the average score for the larger group of students who have taken earth science over the past 20 years. Because he has kept good records, he knows the means and standard deviations for both his group of 24 students and the larger group of 2000 past enrollees. Here are the data he has, and the values for which you will need to compute the z-test.

	Size	Mean	Standard Deviation
Sample	24	100	4.5
Population	2000	99	2.2

Answer 1

 

Dr. McDonald thinks that his group of earth science students is particularly special (in a good way), and he is interested in knowing if their class average falls within the boundaries of the average score for the larger group of students who have taken earth science over the past 20 years. Because he has kept good records, he knows the means and standard deviations for both his group of 24 students and the larger group of 2000 past enrollees. Here are the data he has, and the values for which you will need to compute the z-test.

	Size	Mean	Standard Deviation
Sample	24	100	4.5
Population	2000	99	2.2

Population Standard Deviation = = 2.2

Population Mean = = 99

Sample Mean = = 100

Sample Size = n = 24

1. State the null and research hypotheses:

The null hypothesis states that sample average is equal to the population average. In this problem the null is not rejected , it means that the sample is representative of the population.

:

Hypothesis are :

: = 99

: 99

2.Set the level of risk :

The level of risk is 0.05. This is totally decision of the researcher.

Critical value = z_0.05 = 1.96

3.Set appropriate test statistic:

The appropriate test for the process is one-sample Z test.

4. Compute the test statistic value:

Find standard error of the mean:

SEM = / sqrt(n)

SEM = 2.2 / sqrt(24)

SEM = 0.449

The test statistic = z = (-) / SEM

z = (100-99) / 0.449

z = 2.227 2.23

5.Compare the obtained value and the critical value:

The z value is 2.23 . So , far a test of this null hypothesis at the 0.05 level with 24 participants,

The critical value is 1.96 .

This value represents the value at which chance is the most attractive explanation of why the samples mean and the population mean differ.