4). Measurements of the sodium content in samples of two brands of chocolate bar

Question

4). Measurements of the sodium content in samples of two brands of chocolate bar yield the following results in grams: Brand A 34.36 31.26 37.36 28.52 33.14 32.74 34.34 34.33 30.95 Brand B 41.08 38.22 39.59 38.82 36.24 37.73 35.03 39.22 34.13 34.33 34.98 29.64 40.60 a). In the next part of this question, you will need to find a 98% confidence interval for the difference between the mean sodium contents of the two brands. Before you do it, state which assumption on the variances (equal or different) is more appropriate for the confidence interval and explain why. (3 points) b). According to what you answered part a, find a 98% confidence interval for the differ- c). Write a conclusion about the obtained confidence interval. Is there an actual differ- d). Construct a 98% confidence interval for the difference between the mean sodium ence between the mean sodium contents of the two brands (9 points) ence between the mean sodium contents of the two brands? (3 points) contents of the two brands now using the assumption that you did not choose in part b. (9 points) e). Compare the confidence intervals from parts b and d, which one is wider? why? (1 point)

Explanation / Answer

PART A.
equal variance is more appropriate for the confidence interval.
since we get the higher margin of error value

PARTB.
DIRECT METHOD
given that,
mean(x)=33
standard deviation , s.d1=2.5449
sample size, n1=9
y(mean)=36.8931
standard deviation, s.d2 =3.2211
sample size,n2 =13
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 = [( 33-36.8931) ± t a/2 * sqrt((6.477/9)+(10.375/13)]
= [ (-3.893) ± t a/2 * 1.232]
= [-7.461 , -0.325]
-----------------------------------------------------------------------------------------------

PART C.

interpretations:
1. we are 98% sure that the interval [-7.461 , -0.325] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion

PART D.
given that,
mean(x)=33
standard deviation , s.d1=2.5449
number(n1)=9
y(mean)=36.8931
standard deviation, s.d2 =3.2211
number(n2)=13
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (8*6.477 + 12*10.375) / (22- 2 )
s^2 = 8.816
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 8.816 * (1/9+1/13) )
=1.288
III.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.02
from standard normal table, two tailed and value of |t | with (n1+n2-2) i.e 20 d.f is 2.528
margin of error = 2.528 * 1.288
= 3.255
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (33-36.8931) ± 3.255 ]
= [-7.148 , -0.638]

PART E.
assuming equal is wider, since it has larger t test value with higher degrees of freedom

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.