Question

1. Suppose the amount of money UCLA students spend on movies during a one month period observes normal distribution. A sample is taken containing monthly movie spending in dollars for several UCLA students as 66.72, 50.23, 40.57, 45.53, 60.45, 70.85, 57.49, and 53.46. Round your numbers to two decimal places. All the calculation should be preceded with the formula used.

a. Calculate the sample mean, sample standard deviation, and standard error.

                                   

                       

                       

b. Estimate the average monthly movie spending by all UCLA students with a 95% confidence interval.

           

                       

c. From this sample, can we conclude that the average monthly movie spending by UCLA students is lower than 63.45 dollars at the 0.01 level of significance? (Show 7 steps)

           

d. Suppose the population standard deviation is known with a value of 7.14. Would the conclusions in b and c change? If yes, what would be the new conclusions? ? (Show 7 steps)

                       

Answer 1

1.

TRADITIONAL METHOD
given that,
sample mean, x =55.6625
standard deviation, s =9.643
sample size, n =8
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 9.643/ sqrt ( 8) )
= 3.409
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 7 d.f is 2.365
margin of error = 2.365 * 3.409
= 8.063
III.
CI = x ± margin of error
confidence interval = [ 55.6625 ± 8.063 ]
= [ 47.599 , 63.726 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =55.6625
standard deviation, s =9.643
sample size, n =8
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 7 d.f is 2.365
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 55.6625 ± t a/2 ( 9.643/ Sqrt ( 8) ]
= [ 55.6625-(2.365 * 3.409) , 55.6625+(2.365 * 3.409) ]
= [ 47.599 , 63.726 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 47.599 , 63.726 ] 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
a.
sample mean, x =55.6625
standard deviation, s =9.643
sample size, n =8
standard error = 3.409
b.
95% sure that the interval [ 47.599 , 63.726 ]
c.
Given that,
population mean(u)=63.45
sample mean, x =55.6625
standard deviation, s =9.643
number (n)=8
null, Ho: μ=63.45
alternate, H1: μ<63.45
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =2.998
since our test is left-tailed
reject Ho, if to < -2.998
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =55.6625-63.45/(9.643/sqrt(8))
to =-2.284
| to | =2.284
critical value
the value of |t α| with n-1 = 7 d.f is 2.998
we got |to| =2.284 & | t α | =2.998
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -2.2842 ) = 0.02814
hence value of p0.01 < 0.02814,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=63.45
alternate, H1: μ<63.45
test statistic: -2.284
critical value: -2.998
decision: do not reject Ho
p-value: 0.02814
we do not have enough evidence to support the claim that the average monthly movie spending by UCLA students is lower than 63.45 dollars
d.
population standard deviation =7.14
i.
TRADITIONAL METHOD
given that,
standard deviation, σ =7.14
sample mean, x =55.6625
population size (n)=8
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 7.14/ sqrt ( 8) )
= 2.524
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 * 2.524
= 4.948
III.
CI = x ± margin of error
confidence interval = [ 55.6625 ± 4.948 ]
= [ 50.715,60.61 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =7.14
sample mean, x =55.6625
population size (n)=8
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 = [ 55.6625 ± Z a/2 ( 7.14/ Sqrt ( 8) ) ]
= [ 55.6625 - 1.96 * (2.524) , 55.6625 + 1.96 * (2.524) ]
= [ 50.715,60.61 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [50.715 , 60.61 ] 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
ii.
Given that,
population mean(u)=63.45
standard deviation, σ =7.14
sample mean, x =55.6625
number (n)=8
null, Ho: μ=63.45
alternate, H1: μ<63.45
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 55.6625-63.45/(7.14/sqrt(8)
zo = -3.085
| zo | = 3.085
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =3.085 & | z α | = 2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -3.085 ) = 0.001
hence value of p0.01 > 0.001, here we reject Ho
ANSWERS
---------------
null, Ho: μ=63.45
alternate, H1: μ<63.45
test statistic: -3.085
critical value: -2.326
decision: reject Ho
p-value: 0.001
we have enough evidence to support the claim that the average monthly movie spending by UCLA students is lower than 63.45 dollars.