Question

The researchers suggest that there are occupational differences in mean testosterone level. Medical doctors and university professors are two of the occupational groups for which means and standard deviations are recorded and listed in the following table.

Group	Sample size	Mean	StDev
MD	n1=6n1=6	x¯1=11.21x¯1=11.21	s1=3.73s1=3.73
Prof	n2=5n2=5	x¯2=11.6x¯2=11.6	s2=2.14s2=2.14

Let us denote:

• μ1:μ1: population mean testosterone among medical doctors,
• μ2:μ2: population mean testosterone among university professors,
• σ1:σ1: population standard deviation of testosterone among medical doctors,
• σ2:σ2: population standard deviation of testosterone among university professors.

Case 1: Assume that the population standard deviations are unequal, i.e. σ1≠σ2σ1≠σ2.
What is the standard error of the difference in sample mean x¯1−x¯2x¯1−x¯2? i.e. s.e.(x¯1−x¯2)=s.e.(x¯1−x¯2)= [answer to 4 decimal places]

Tries 0/5

Which of the following options gives the formula for 95% confidence interval for μ1−μ2μ1−μ2?
−0.39∓1.86×s.e.(x¯1−x¯2)−0.39∓1.86×s.e.(x¯1−x¯2)
−0.39∓3.36×s.e.(x¯1−x¯2)−0.39∓3.36×s.e.(x¯1−x¯2)
−0.39∓2.9×s.e.(x¯1−x¯2)−0.39∓2.9×s.e.(x¯1−x¯2)
−0.39∓2.31×s.e.(x¯1−x¯2)−0.39∓2.31×s.e.(x¯1−x¯2)
−0.39∓1.4×s.e.(x¯1−x¯2)−0.39∓1.4×s.e.(x¯1−x¯2)

Tries 0/3

Are there significant difference between mean testosterone levels of medical doctors and university professors?
no, because x¯1=x¯2x¯1=x¯2
yes, because x¯1≠x¯2x¯1≠x¯2
yes, because the entire 95% confidence interval for μ1−μ2μ1−μ2 does not contain 0
no, because the 95% confidence interval for μ1−μ2μ1−μ2 contains 0

Tries 0/3

Case 2: Now assume that the population standard deviations are equal, i.e. σ1=σ2σ1=σ2.
Compute the pooled standard deviation, spooledspooled [answer to 4 decimal places]

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Which of the following options gives the formula for 95% confidence interval for μ1−μ2μ1−μ2 for pooled situation?
−0.39∓1.38×1.8922−0.39∓1.38×1.8922
−0.39∓3.25×1.8922−0.39∓3.25×1.8922
−0.39∓2.26×1.8922−0.39∓2.26×1.8922
−0.39∓1.83×1.8922−0.39∓1.83×1.8922
−0.39∓2.82×1.8922−0.39∓2.82×1.8922

Tries 0/3

What is the margin of error of the 95% pooled confidence interval of μ1−μ2μ1−μ2? [answer to 4 decimal places]

Tries 0/5

Answer 1

case I

Sample #1   ---->        
mean of sample 1,    x̅1=   11.21          
standard deviation of sample 1,   s1 =    3.73          
size of sample 1,    n1=   6          
                  
Sample #2   ---->     
mean of sample 2,    x̅2=   11.600          
standard deviation of sample 2,   s2 =    2.14          
size of sample 2,    n2=   5          
                  
difference in sample means = x̅1-x̅2 =    11.210   -   11.6000   =   -0.3900
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1.7985          
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t-critical value =    t α/2 =    2.31

confidence interval given by , difference of means ± critical value*s.e

−0.39 ∓ 2.31×s.e.(x¯1−x¯2)

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no, because the 95% confidence interval for μ1−μ2 contains 0

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Case II

pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    3.1249
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Degree of freedom, DF=   n1+n2-2 =    9
t-critical value =    t α/2 =    2.2622
std error , SE =    Sp*√(1/n1+1/n2) =    1.8922  

confidence interval is

−0.39∓2.26×1.8922

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margin of error, E = t*SE =    2.2622   *   1.8922   =   4.2804