Use the following scenario and data for questions 35 to 39: An market analyst is

Question

Use the following scenario and data for questions 35 to 39: An market analyst is interested in the expenses on toys by parents for their kids living in rural or urban areas Random samples of n 54 families living in rural areas and m 38 families living in urban areas are selected. The sample means and standard deviations on annual expenses on toys are found to be -$88.85, -S94.80 , S-$13.10 and s2-$12.42. The two populations are assumed to have equal variances although their population variances are unknown. You are required to construct a confidence interval for the differences in the two population means, Ah-A-with a confidence coefficient l-a-09. (10) Note: The following solutions were obtained by pooling the sample variances or standard deviations because I taught the material that way in the Spring semester of 2015. You do not pool the variances or standard deviations when the sample sizes are large for your quizzes and tests because this is the way we learned this semester. Because we are not pooling the variances, the answers will not match the choices for the following questions. You may use Quiz 22 as a good reference for problems of this type. 35·The pooled standard deviation sp is closest to (3) (A) 13.1000 (B) 12.4200 (C) 12.8248 (D) 164.4757 (E) none of the above 36. The standard deviation of- ,Le., s-h' s closest to (3) (A) 3.5812 (B) 1.6450 (C) 12.8248 (D) 2.7155 (E) none of the above 37. The left limit L of this confidence interval is closest to (3) (A)-10.4171 (B)-5.9500 (C)-1.4829 (D)5.9500 (E) none of the above

Explanation / Answer

TRADITIONAL METHOD
given that,
mean(x)=88.85
standard deviation , s.d1=13.1
number(n1)=54
y(mean)=94.8
standard deviation, s.d2 =12.42
number(n2)=38
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (53*171.61 + 37*154.2564) / (92- 2 )
s^2 = 164.4757
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 164.4757 * (1/54+1/38) )
=2.7155
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.1
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 90 d.f is 1.662
margin of error = 1.662 * 2.7155
= 4.5132
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (88.85-94.8) ± 4.5132 ]
= [-10.4632 , -1.4368]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=88.85
standard deviation , s.d1=13.1
sample size, n1=54
y(mean)=94.8
standard deviation, s.d2 =12.42
sample size,n2 =38
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 88.85-94.8) ± t a/2 * sqrt( 164.4757 * (1/54+1/38) ]
= [ (-5.95) ± 4.5132 ]
= [-10.4632 , -1.4368]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-10.4632 , -1.4368]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

35) 164.4757 36) 2.7155 37) -10.4171 38) -1.4829 39) with 90% COnfidence -10.4171<=u1-u2<=-1.4829 39

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