In: Statistics and Probability
Are medical students more motivated than law students? A randomly selected group of each were administered a survey of attitudes toward Life, which measures motivation for upward mobility. The scores are summarized below. 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 |
---|---|---|---|
Medical | n1=4n1=4 | x¯1=87.01x¯1=87.01 | s1=6.7s1=6.7 |
Law | n2=10n2=10 | x¯2=85.95x¯2=85.95 | s2=16.5s2=16.5 |
Let us denote:
If the researcher is interested to know whether the mean
testosterone level among medical doctors is higher than that among
university professors, what are the appropriate hypotheses he
should test?
H0:μ1=μ2H0:μ1=μ2 against
Ha:μ1≠μ2Ha:μ1≠μ2.
H0:x¯1=x¯2H0:x¯1=x¯2 against
Ha:x¯1>x¯2Ha:x¯1>x¯2.
H0:x¯1=x¯2H0:x¯1=x¯2 against
Ha:x¯1<x¯2Ha:x¯1<x¯2.
H0:μ1=μ2H0:μ1=μ2 against
Ha:μ1>μ2Ha:μ1>μ2.
H0:x¯1=x¯2H0:x¯1=x¯2 against
Ha:x¯1≠x¯2Ha:x¯1≠x¯2.
H0:μ1=μ2H0:μ1=μ2 against
Ha:μ1<μ2Ha:μ1<μ2.
Tries 0/3 |
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 |
Rejection region: We reject H0H0 at 1% level of significance
if:
t<−3.05t<−3.05.
t>3.05t>3.05.
t<−2.68t<−2.68.
t>2.68t>2.68.
|t|>3.05|t|>3.05.
None of the above.
Tries 0/3 |
The value of the test-statistic is: Answer to 3 decimal places.
Tries 0/5 |
If α=0.01α=0.01, and the p-value is 0.4335, what will be your
conclusion?
There is not enough information to conclude.
Do not reject H0H0.
Reject H0H0.
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]
Tries 0/5 |
Rejection region: We reject H0H0 at 1% level of significance
if:
t>3.05t>3.05.
t<−2.68t<−2.68.
t>2.68t>2.68.
|t|>3.05|t|>3.05.
t<−3.05t<−3.05.
None of the above.
Tries 0/3 |
The value of the test-statistic is: Answer to 3 decimal places.
Tries 0/5 |
If α=0.01α=0.01, , and the p-value is 0.4525, what will be your
conclusion?
Reject H0H0.
Do not reject H0H0.
There is not enough information to conclude.
Tries 0/3 |
H0:μ1=μ2 against Ha:μ1>μ2
--------------
Sample #1 ---->
mean of sample 1, x̅1= 87.01
standard deviation of sample 1, s1 =
6.7
size of sample 1, n1= 4
Sample #2 ---->
mean of sample 2, x̅2= 85.950
standard deviation of sample 2, s2 =
16.50
size of sample 2, n2= 10
difference in sample means = x̅1-x̅2 =
87.010 - 85.9500 =
1.0600
std error , SE = √(s1²/n1+s2²/n2) =
6.2006
----------------------------------
α=0.01
t>2.68
---------------------
t-statistic = ((x̅1-x̅2)-µd)/SE = ( 1.0600 / 6.2006 ) = 0.1710
---------------
p-value>α , Do not reject null hypothesis
There is not enough information to conclude that the mean testosterone level among medical doctors is higher than that among university professors
================================================================
case II
ASSUMING equal variance |
Degree of freedom, DF= n1+n2-2 =
12
t-critical value , t* =
2.6810
t>2.68
difference in sample means = x̅1-x̅2 =
87.0100 - 86.0 =
1.06
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 14.68
std error , SE = Sp*√(1/n1+1/n2) =
8.6829
t-statistic = ((x̅1-x̅2)-µd)/SE = ( 1.0600
- 0 ) / 8.68
= 0.122
0.4524
p-value>α , Do not reject null hypothesis
There is not enough information to conclude that the mean
testosterone level among medical doctors is higher than that among
university professors