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Ch. 24 notes: Comparing 2 means: 2 sample t-interval and test
2-sample T-test
HYPOTHESES:
Ho: 1  2

 
Ha: 1  2
 

 
CONDITIONS:
Just double the 1- sample conditions…
1. 2 independent SRS
2. 10 % condition for both
pop1  10n1
pop2  10n2
3. 2 Normal Populations OR n1  30 & n2  30
Conditions met  t- distribution  2-sample t- Test
MECHANICS:
t
statistic  parameter

std . dev. of statistic ( SE )
x1  x2
 s1 
2
n1
s 
 2
2
n2
P-Value: P(t <, > test statistic l df = #)
To find the degrees of freedom –
On the calculator – round to the nearest whole number
OR
Use the smaller degrees of freedom of the two samples
CONCLUSION:
Same two sentences (includes the word “difference”)
-
We reject/ fail to reject ….
We have sufficient/insufficient evidence that the true difference in means is ….(use Ha)
Example: Resting pulse rates for a random sample of 26 smokers had a mean of 80 beats per minute (bpm) and a
standard deviation of 5 bpm. Among 32 randomly chosen nonsmokers, the mean and standard deviation were 74
and 6 bpm. Both sets of data were roughly symmetric and had no outliers. Is there evidence of a difference in
mean pulse rate between smokers and non-smokers? How big?
xS  80
sS  5
nS  26
xN  74
sN  6
nN  32
Ho: S   N
Ha: S   N
Conditions:
1. 2 independent SRS
1. Smokers and Non-smokers are independent
of one another
Both samples stated SRS
2. 10 % condition for both
2. There are more than 260 smokers and
more than 320 non-smokers so we are
sampling less than 10% of the populations
3. 2 Normal Populations OR n1  30 & n2  30
3. Both data sets were roughly symmetric
with no outliers so we can assume both are
nearly normal
Conditions met  t- distribution  2-sample t- Test
t
80  74
 5
2
26
 6

2
 4.154
32
2* P(t > 4.154 l df = 56) = 1.129*10 4
We reject the Ho because the P-Value of 1.129*10 4 <   .05 .
We have sufficient evidence that the true average pulse rate (beats per minute) between
smokers and non-smokers is different.
2-sample t-Interval
CONDITIONS:
Same as 2- sample t-Test
Conditions met  t-distribution  2-sample t-Interval
MECHANICS:
statistic  (critical value)(standard deviation of statistic)

( x1  x2 )  (t ) 


*
 s1 
n1
2
s 
 2
n2
2

  ( a, b)


INVT
Difference of means
2-sample T-Interval (on calculator)
Degrees of Freedom – from calculator
CONCLUSION:
We are ____% confident that the true difference between the mean of ______ & ________ is
between ___a____ and __b____ units.
Example: Since we rejected our Ho, complete the interval for our example on smokers above.
Conditions met  t- distribution  2- sample t- Interval

(80  74)  (2.002) 


 5
2
26
df = 56

  (3.106, 8.894)
32 

 6

2
We are 95% confident that the true difference between the average resting heart rate of smokers and nonsmokers is between 3.106 bpm and 8.894 bpm.
Example 2: Here are the saturated fat content (in grams) for several pizzas sold by two national chains. Do the
two pizza chains have significantly different mean saturated fat contents?
17
8
6
Brand PJ
11
Brand D
12
12
7
3
10
15
11
4
Hypotheses:
Ho : D  PJ
Ha : D  PJ
8
7
9
5
8
11
4
8
10
11
4
5
10 5 16 16
13 13 11 12
7 9
5
Conclusion:
We reject the Ho because the P-Value of .0000315 <   .05 .
We have sufficient evidence that the average saturated fat content of brand D is not equal to brand PJ.
Book Problems: p. 579 #2, 3, 11, 14, 27, 36