The IQ score for a random sample of 88 people has a sample m

The IQ score for a random sample of 88 people has a sample mean of 102 and sample standard deviation 0f 14.

a. Give a 95% confidence interval for the population mean IQ.

b. The claim is made that the population mean IQ exceeds 100. Perform an appropriate hypothesis test at a significance level of 5%. State clearly your conclusions.

Solution

a)

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    102          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    14          
n = sample size =    88          
              
Thus,              
              
Lower bound =    99.07493992          
Upper bound =    104.9250601          
              
Thus, the confidence interval is              
              
(   99.07493992   ,   104.9250601   ) [ANSWER]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   100  
Ha:    u   >   100  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    +   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    102          
uo = hypothesized mean =    100          
n = sample size =    88          
s = standard deviation =    14          
              
Thus, z = (X - uo) * sqrt(n) / s =    1.340118789          
              
Also, the p value is              
              
p =    0.090103364          
              
Comparing z and zcrit (or, p > 0.05), we   FAIL TO REJECT THE NULL HYPOTHESIS.  
      
Thus, there is no significant evidence that the population mean IQ exceeds 100. [CONCLUSION]