Question

7) A food company is developing a new granola bar, and its market analysts are currently working on preliminary studies of the packaging design. To help with a marketing strategy, the company was first interested in whether the appeal of the packaging design for the new product (and hence the appeal of the product itself) would be different for each gender. There were 100 male and 100 female volunteers available for purposes of evaluation. Both genders rated the design on a scale of 1 to 10, 1 being "very unappealing" and 10 being "very appealing." The data summaries follow. n x-bar s Females 100 8.0 2.0 Males 100 7.4 1.5 Let mu1 and mu2 represent the mean ratings we would observe for the populations of females and males, respectively, and assume our samples can be regarded as samples from these populations. You may assume the 2-sample t-procedures are safe to use. Does the data given evidence of a difference in appeal for females and males? To explore the suspicion above, conduct a significance test. Calculator output of the test is given below. a) The null hypothesis is μ1__μ2 Fill in the blank as appropriate. b) Give the test decision: (reject or do not reject) c) Evidence _______________(favors or does not favor) that there is a difference in the mean appeal ratings of females and males. d) A 95% confidence interval for the difference in the mean appeal rating between females and males is 0.6±0.493 e) Interpret this by filling in the blanks as appropriate: 95% confident that the true mean difference in appeal rating between females and males is between ______________ and ___________________ (give three decimals for bounds). This supports the decision of test because 0 ____________(is or is not) between the bounds.

Answer 1

7.

Given that,
mean(x)=8
standard deviation , s.d1=2
number(n1)=100
y(mean)=7.4
standard deviation, s.d2 =1.5
number(n2)=100
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =1.984
since our test is two-tailed
reject Ho, if to < -1.984 OR if to > 1.984
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =8-7.4/sqrt((4/100)+(2.25/100))
to =2.4
| to | =2.4
critical value
the value of |t α| with min (n1-1, n2-1) i.e 99 d.f is 1.984
we got |to| = 2.4 & | t α | = 1.984
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.4 ) = 0.018
hence value of p0.05 > 0.018,here we reject Ho
ANSWERS
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a.
null, Ho: u1 = u2
alternate, H1: u1 != u2
b.
test statistic: 2.4
critical value: -1.984 , 1.984
decision: reject Ho
p-value: 0.018
c.
we have enough evidence to support the claim that there is a difference in the mean appeal ratings of females and males.
d.
TRADITIONAL METHOD
given that,
mean(x)=8
standard deviation , s.d1=2
number(n1)=100
y(mean)=7.4
standard deviation, s.d2 =1.5
number(n2)=100
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((4/100)+(2.25/100))
= 0.25
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 99 d.f is 1.984
margin of error = 1.984 * 0.25
= 0.496
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (8-7.4) ± 0.496 ]
= [0.104 , 1.096]
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DIRECT METHOD
given that,
mean(x)=8
standard deviation , s.d1=2
sample size, n1=100
y(mean)=7.4
standard deviation, s.d2 =1.5
sample size,n2 =100
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 = [( 8-7.4) ± t a/2 * sqrt((4/100)+(2.25/100)]
= [ (0.6) ± t a/2 * 0.25]
= [0.104 , 1.096]
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interpretations:
1. we are 95% sure that the interval [0.104 , 1.096] 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
95% sure that the interval [0.104 , 1.096] in this context zero is included in this interval.