Question

Question1

A university lecturer is interested in comparing the engagement levels of first-year statistics students. In a previous nation-wide study, engagement levels of all university students were found to be normally distributed, with µ=60.00. The lecturer collects a random sample of 50 first-year students and the following statistics are obtained: M=65.43, SD=7.82.

What statistical procedure should be used, to test whether there is a significant mean difference in engagement levels between the lecturer’s first year students and the population average?

	a.	One sample Z-test.
	b.	Dependent samples t-test.
	c.	One sample t-test.
	d.	Independent samples t-test.

Question 2

A university lecturer is interested in comparing the enthusiasm levels of first-year statistics students. In a previous nation-wide study, enthusiasm levels were found to be normally distributed, with µ=70.00, σ=5.00. The lecturer collects a convenience sample of 50 first-year students and finds that her students have a mean enthusiasm level equal to 65.24.

What statistical procedure should be used, to test whether there is a significant mean difference in enthusiasm levels between the lecturer’s first year students and the population average?

	a.	Two sample Z-test
	b.	One sample Z-test.
	c.	Independent samples t-test.
	d.	One sample t-test.

Question 3

An organisational psychologist hypothesised that employee IQ levels of major Australian banks differ significantly to the general population. To test this, he performed a Z-test. Listed below are the IQ scores of 20 random employees:

105, 98, 103, 115,116,118,121,132,95,105,108,132,114,118,126,127,127,124,119,138.

If IQ scores are normally distributed, with µ=100, σ=15, what is the Z-statistic? Use these figures to calculate and select the correct the Z-statistic below.

	a.	17.05
	b.	3.35
	c.	1.14
	d.	5.08

Answer 1

1)

One sample t-test. because σ is unknown

2)

One sample Z-test because σ is known

3)

population std dev ,    σ =    15.0000                  
Sample Size ,   n =    20                  
Sample Mean,    x̅ = ΣX/n =    117.0500                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   15.0000   / √    20   =   3.3541      
Z-test statistic= (x̅ - µ )/SE = (   117.050   -   100   ) /    3.3541   =   5.08 (answer)