4. (21.57 S-AQ) Healthy women aged 18 to 40 participated in a study of eating ha

Question

4. (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¯ ss

Unrestrained 9 63.83 24.72

Restrained 11 34 39.8

The critical value, t*, from the t distribution for confidence interval 80 % using the conservative degrees of freedom is:

t* = ? . (DO NOt ROUND)

The margin of error, E, for this confidence interval is ? (Round to 2 decimal places.)

Give a 80 % confidence interval that describes the effect of restraint:

?  to ? . (Round to 2 decimal places.)

Explanation / Answer

4.

TRADITIONAL METHOD
given that,
mean(x)=63.83
standard deviation , s.d1=24.72
number(n1)=9
y(mean)=34
standard deviation, s.d2 =39.8
number(n2)=11
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((611/9)+(1584/11))
= 15
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 8 d.f is 1
margin of error = 1.4 * 15
= 20
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (63.83-34) ± 20 ]
= [9 , 50]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=63.83
standard deviation , s.d1=24.72
sample size, n1=9
y(mean)=34
standard deviation, s.d2 =39.8
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 = [( 63.83-34) ± t a/2 * sqrt((611/9)+(1584/11)]
= [ (30) ± t a/2 * 15]
= [9 , 50]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 80% sure that the interval [9 , 50] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 80% of these intervals will contains the true population proportion

II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 8 d.f is 1.4
margin of error = 1.4 * 14.56
= 20.38

III)
DIRECT METHOD
given that,
mean(x)=63.83
standard deviation , s.d1=24.72
sample size, n1=9
y(mean)=34
standard deviation, s.d2 =39.8
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 = [( 63.83-34) ± t a/2 * sqrt((611.08/9)+(1584.04/11)]
= [ (29.83) ± t a/2 * 14.56]
= [9.45 , 50.21]
80 % confidence interval that the effect of restraint it will slight change in the confidence interval lower and upper bound.

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