An economics professor randomly selected 100 millionaires in the U. S. The average age of these millionaires was 52.1 years with a standard deviation of 12.3 years. What is a 95% confidence interval for the mean age, μ, of all U. S. millionaires?

The 95% confidence interval would be given by (49.66;54.54)    

Step-by-step explanation:

Information given

represent the sample mean

population mean (variable of interest)

s=12.3 represent the sample standard deviation

n=100 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

  (1)

The degrees of freedom are given by:

The Confidence is 0.95 or 95%, and the significance would be and , the critical value for this case is:

Replacing the info given we got:

The 95% confidence interval would be given by (49.66;54.54)    

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1  

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