Question

You wish to test the claim that the average IQ score is less than 100 at the .025 significance level. You determine the hypotheses are:

Ho: μ=100

H1:μ<100

You take a simple random sample of 96 individuals and find the mean IQ score is 96.2, with a standard deviation of 15.7. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is not known and once with assuming that it is known.

Round to three decimal places where appropriate.

Assume Population Standard Deviation is NOT known	Assume Population Standard Deviation is 15
Test Statistic: t =	Test Statistic: z =
Critical Value: t =	Critical Value: z =
p-value:	p-value:
Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis	Conclusion About the Null: Reject the null hypothesis Fail to reject the null hypothesis
Conclusion About the Claim: There is sufficient evidence to support the claim that the average IQ score is less than 100. There is NOT sufficient evidence to support the claim that the average IQ score is less than 100. There is sufficient evidence to warrant rejection of the claim that the average IQ score is less than 100. There is NOT sufficient evidence to warrant rejection of the claim that the average IQ score is less than 100.	Conclusion About the Claim: There is sufficient evidence to support the claim that the average IQ score is less than 100. There is NOT sufficient evidence to support the claim that the average IQ score is less than 100. There is sufficient evidence to warrant rejection of the claim that the average IQ score is less than 100. There is NOT sufficient evidence to warrant rejection of the claim that the average IQ score is less than 100.

Is there a significant difference between when we know the population standard deviation and when we don't? Explain.

Answer 1

Assume Population Standard Deviation is NOT known

Standard Error , SE = s/√n =   15.7/√96=   1.6024
t-test statistic= (x̅ - µ )/SE =    (96.2-100)/1.6024=   -2.3715

critical t value, t* =        -1.9853
p-Value   =   0.0099

Conclusion About the Null:

• Reject the null hypothesis

There is sufficient evidence to support the claim that the average IQ score is less than 100.

=================================

Assume Population Standard Deviation is 15

Standard Error , SE = σ/√n =   15/√96=   1.5309
Z-test statistic= (x̅ - µ )/SE =    (96.2-100)/1.5309=   -2.4821
      
critical z value, z* =       -1.9600
      
p-Value   =   0.0065
Decision:   p-value≤α, Reject null hypothesis   
There is sufficient evidence to support the claim that the average IQ score is less than 100.