Question

(21.57 S-AQ) Healthy women aged 18 to 40 participated in a study of eating habits. Subjects were given bags of potato chips and bottled water and invited to snack freely. Interviews showed that some women were trying to restrain their diet out of concern about their weight. How much effect did these good intentions have on their eating habits? Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation):

Group	n	x¯¯¯x¯	SEM
Unrestrained	9	63.86	5.6
Restrained	11	32	11

The standard error of the difference of sample means (±0.0001) is: .

The critical value (±0.001) from the t distribution for confidence interval 95% using the conservative degrees of freedom is:   
    t* = .

Give a 95% confidence interval (±0.01) that describes the effect of restraint:
  to    .

Based on this interval, is there a significant difference between the two groups?

No

Yes

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=63.86
standard deviation , s.d1=16.8
number(n1)=9
y(mean)=32
standard deviation, s.d2 =36.482
number(n2)=11
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((282.24/9)+(1330.936/11))
= 12.343
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 2.306
margin of error = 2.306 * 12.343
= 28.463
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (63.86-32) ± 28.463 ]
= [3.397 , 60.323]
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DIRECT METHOD
given that,
mean(x)=63.86
standard deviation , s.d1=16.8
sample size, n1=9
y(mean)=32
standard deviation, s.d2 =36.482
sample size,n2 =11
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 63.86-32) ± t a/2 * sqrt((282.24/9)+(1330.936/11)]
= [ (31.86) ± t a/2 * 12.343]
= [3.397 , 60.323]
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interpretations:
1. we are 95% sure that the interval [3.397 , 60.323] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion