A food company is concerned about recent criticism of the sugar content of their

Question

A food company is concerned about recent criticism of the sugar content of their children's cereals. The company would like to know if there is a difference between the sugar content of children's cereals and adults' cereals. The data show the sugar content (as a percentage of weight) of several national brands of children's and adults' cereals. Complete parts a through d.

Children's cereal:

43.9, 55.7, 46.9, 40.2, 51.2, 46, 54.3, 41.3, 43.1, 44.6, 46.6, 43.3, 36.8, 57, 49.1, 52.7, 35.6, 55.5, 42.8, 34.5

Adults' cereal:

21, 26.6, 2.9, 7.4, 3.4, 22, 19, 13.2, 24.6, 6.5, 6, 10.8, 18, 14.4, 3.3, 17.8, 0.8, 1.4, 0.5, 7.1, 12, 3.9, 3, 0.8, 8.5,1.8, 18.4, 8.2, 19.5, 13.5

a) Write the null and alternative hypotheses.

Let group C correspond to children's cereals and group A correspond to adults' cereals. Choose the correct answer below.

Answer :

H0: C A =0

HA: C A 0

b) Check the conditions.

Which of the following conditions are satisfied for the given data? Select all that apply.

Answers:

Independent groups assumption

Nearly normal condition

Randomization condition

c) Find the 95% confidence interval for the positive difference in means.

The 95% confidence interval is

( grams, grams).

d) Is there a significant difference in mean sugar content between these two types of cereals? Explain.

Examine the 95% confidence interval for the difference of means found in the previous step. Note that the interval does not contain 0.

Use this information to determine if there is a significant difference in mean sugar content between these two types of cereals.

Explanation / Answer

PART A & B.
Given that,
mean(x)=46.055
standard deviation , s.d1=6.7667
number(n1)=20
y(mean)=10.5433
standard deviation, s.d2 =7.9497
number(n2)=30
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =46.055-10.5433/sqrt((45.78823/20)+(63.19773/30))
to =16.9372
| to | =16.9372
critical value
the value of |t | with min (n1-1, n2-1) i.e 19 d.f is 2.093
we got |to| = 16.93723 & | t | = 2.093
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 16.9372 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, H0: C A =0
alternate, HA: C A 0
Independent groups assumption

PART C.
TRADITIONAL METHOD
given that,
mean(x)=46.055
standard deviation , s.d1=6.7667
number(n1)=20
y(mean)=10.5433
standard deviation, s.d2 =7.9497
number(n2)=30
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((45.788/20)+(63.198/30))
= 2.097
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 19 d.f is 2.093
margin of error = 2.093 * 2.097
= 4.388
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (46.055-10.5433) ± 4.388 ]
= [31.123 , 39.9]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=46.055
standard deviation , s.d1=6.7667
sample size, n1=20
y(mean)=10.5433
standard deviation, s.d2 =7.9497
sample size,n2 =30
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 = [( 46.055-10.5433) ± t a/2 * sqrt((45.788/20)+(63.198/30)]
= [ (35.512) ± t a/2 * 2.097]
= [31.123 , 39.9]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [31.123 , 39.9] 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

we claim that significant difference in mean sugar content between these two types of cereals

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