Question

You wish to test the claim that the average IQ score is less than 100 at the .05 significance level. You determine the hypotheses are:

Ho: μ=100

H1:μ<100

You take a simple random sample of 38 individuals and find the mean IQ score is 95.8, with a standard deviation of 15.2. 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

Assuming the population standard deviation is not known , then t-test will be used

Level of Significance ,    α =    0.05
sample std dev ,    s =    15.200
Sample Size ,   n =    38
Sample Mean,    x̅ =   95.8
      
degree of freedom=   DF=n-1=   37
      
Standard Error , SE =   s/√n =   2.4658
      
t-test statistic=   (x̅ - µ )/SE =    -1.703

critical t value, t*   =   -1.6871   [Excel formula =t.inv(α,df) ]
          
p-Value   =   0.0484
Conclusion:     p-value<α, Reject null hypothesis       

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

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assuming the population standard deviation is known and is equal to 15,Z test will be used

Level of Significance ,    α =    0.05  
population std dev ,    σ =    15.000  
Sample Size ,   n =    38  
Sample Mean,    x̅ =   95.8  
          
'   '   '  
          
Standard Error , SE =   σ/√n =   2.4333  
          
Z-test statistic=   (x̅ - µ )/SE =    -1.726  
          
critical z value, z*   =   -1.6449   [Excel formula =NORMSINV(α) ]
          
p-Value   =   0.0422   [excel function =NORMSDIST(z)]
Conclusion:     p-value<α, Reject null hypothesis       

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

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when we  know the population standard deviation , then z test is used

and when we dont know, then t test is used

but when sample size is large enough(n≥30), then distribution of population is apprximated as normally distributed

and z test can be used instead of t.