Question

Confidence intervals, effect sizes, and Valentine’s Day spending: According to the Nielsen Company, Americans spend $345 million on chocolate during the week of Valentine’s Day. Let’s assume that we know the average married person spends $45, with a population standard deviation of $16. In February 2009, the U.S. economy was in the throes of a recession. Comparing data for Valentine’s Day spending in 2009 with what is generally expected might give us some indication of the attitudes during the recession. a. Compute the 95% confidence interval for a sample of 18 married people who spent an average of $38. b. How does the 95% confidence interval change if the sample mean is based on 180 people? c. If you were testing a hypothesis that things had changed under the financial circumstances of 2009 as compared to previous years, what conclusion would you draw in part (a) versus part (b)? d. Compute the effect size based on these data and describe the size of the effect.

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =16
sample mean, x =38
population size (n)=18
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 16/ sqrt ( 18) )
= 3.771
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 3.771
= 7.392
III.
CI = x ± margin of error
confidence interval = [ 38 ± 7.392 ]
= [ 30.608,45.392 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =16
sample mean, x =38
population size (n)=18
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 38 ± Z a/2 ( 16/ Sqrt ( 18) ) ]
= [ 38 - 1.96 * (3.771) , 38 + 1.96 * (3.771) ]
= [ 30.608,45.392 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [30.608 , 45.392 ] contains the true population mean
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 mean
b.
TRADITIONAL METHOD
given that,
standard deviation, σ =16
sample mean, x =38
population size (n)=180
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 16/ sqrt ( 180) )
= 1.193
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1.193
= 2.337
III.
CI = x ± margin of error
confidence interval = [ 38 ± 2.337 ]
= [ 35.663,40.337 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =16
sample mean, x =38
population size (n)=180
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 38 ± Z a/2 ( 16/ Sqrt ( 180) ) ]
= [ 38 - 1.96 * (1.193) , 38 + 1.96 * (1.193) ]
= [ 35.663,40.337 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [35.663 , 40.337 ] contains the true population mean
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 mean
c.
Given that,
population mean(u)=45
standard deviation, σ =16
sample mean, x =38
number (n)=18
null, Ho: μ=45
alternate, H1: μ!=45
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 38-45/(16/sqrt(18)
zo = -1.856
| zo | = 1.856
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.856 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.856 ) = 0.063
hence value of p0.05 < 0.063, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=45
alternate, H1: μ!=45
test statistic: -1.856
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.063
we do not have enough evidence to support the claim that things had changed under the financial circumstances of 2009 as compared to previous years

conclusion :
from part (a) and (b)
part (a)
95% sure that the interval [30.608 , 45.392 ]
part(b)
95% sure that the interval [35.663 , 40.337 ]
in both cases sample size is increases from part (a) to (b) then confidence intervals are changed.

d.
Effective size = modulus of (sample mean - population mean )/standard deviation
Effective size =modulus of (38-45)/16
Effective size =modulus of (-0.4375)
Effective size = 0.4375
medium effect