Question

A researcher with the Department of Education followed a cohort of students who graduated from high school in a certain year, monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to determine whether students who first attended community college took longer to attain a bachelor's degree than those who immediately attended and remained at a 4-year institution. The data in the table attached below summarize the results of his study. Complete parts a) through e) below. Click here to view the sample data. Click here to view Student's t-distribution table.

(a) Does the evidence suggest that community college transfer students take longer to attain a​ bachelor's degree? Use an α = 0.05 level of significance. Perform a hypothesis test. Determine the null and alternative hypotheses.

(b) Determine the test statistic (t = ?) and the P-value (P = ?). Round to two decimal places as needed.

(c) Construct a 90​% confidence interval for (μ_{community college} − μ_{no transfer}) to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. The bounds of a ​(1 − α​) •​ 100% confidence interval about μ₁ − μ₂ can be found using the formulas, where t_α/2 is the critical value. Round to six decimal places as needed.

(d) Construct a 95​% confidence interval for (μ_{community college} − μ_{no transfer}) to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. The bounds of a ​(1 − α​) •​ 100% confidence interval about μ₁ − μ₂ can be found using the formulas, where t_α/2 is the critical value. Round to six decimal places as needed.

(e) Construct a 98​% confidence interval for (μ_{community college} − μ_{no transfer}) to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. The bounds of a ​(1 − α​) •​ 100% confidence interval about μ₁ − μ₂ can be found using the formulas, where t_α/2 is the critical value. Round to six decimal places as needed.

Answer 1

Given that,
mean(x)=5.39
standard deviation , s.d1=1.149
number(n1)=255
y(mean)=4.33
standard deviation, s.d2 =1.016
number(n2)=1169
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.651
since our test is right-tailed
reject Ho, if to > 1.651
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =5.39-4.33/sqrt((1.3202/255)+(1.03226/1169))
to =13.616
| to | =13.616
critical value
the value of |t α| with min (n1-1, n2-1) i.e 254 d.f is 1.651
we got |to| = 13.61631 & | t α | = 1.651
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 13.6163 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 13.616
critical value: 1.651
decision: reject Ho
b.
p-value: 0
we have enough evidence to support the claim that whether the student's who first attended community college took longer to attain a bachelor's degree than
those who immediately attended ans remained at a 4 year institution.
c.
TRADITIONAL METHOD
given that,
mean(x)=5.39
standard deviation , s.d1=1.149
number(n1)=255
y(mean)=4.33
standard deviation, s.d2 =1.016
number(n2)=1169
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((1.320201/255)+(1.032256/1169))
= 0.077848
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 254 d.f is 1.650875
margin of error = 1.651 * 0.077848
= 0.128527
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.39-4.33) ± 0.128527 ]
= [0.931473 , 1.188527]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5.39
standard deviation , s.d1=1.149
sample size, n1=255
y(mean)=4.33
standard deviation, s.d2 =1.016
sample size,n2 =1169
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 = [( 5.39-4.33) ± t a/2 * sqrt((1.320201/255)+(1.032256/1169)]
= [ (1.06) ± t a/2 * 0.077848]
= [0.931473 , 1.188527]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [0.931473 , 1.188527] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
d.
TRADITIONAL METHOD
given that,
mean(x)=5.39
standard deviation , s.d1=1.149
number(n1)=255
y(mean)=4.33
standard deviation, s.d2 =1.016
number(n2)=1169
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((1.320201/255)+(1.032256/1169))
= 0.077848
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 254 d.f is 1.969348
margin of error = 1.969 * 0.077848
= 0.153282
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.39-4.33) ± 0.153282 ]
= [0.906718 , 1.213282]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5.39
standard deviation , s.d1=1.149
sample size, n1=255
y(mean)=4.33
standard deviation, s.d2 =1.016
sample size,n2 =1169
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 = [( 5.39-4.33) ± t a/2 * sqrt((1.320201/255)+(1.032256/1169)]
= [ (1.06) ± t a/2 * 0.077848]
= [0.906718 , 1.213282]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [0.906718 , 1.213282] 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
e.
TRADITIONAL METHOD
given that,
mean(x)=5.39
standard deviation , s.d1=1.149
number(n1)=255
y(mean)=4.33
standard deviation, s.d2 =1.016
number(n2)=1169
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((1.320201/255)+(1.032256/1169))
= 0.077848
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.02
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 254 d.f is 2.341118
margin of error = 2.341 * 0.077848
= 0.182242
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.39-4.33) ± 0.182242 ]
= [0.877758 , 1.242242]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5.39
standard deviation , s.d1=1.149
sample size, n1=255
y(mean)=4.33
standard deviation, s.d2 =1.016
sample size,n2 =1169
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 = [( 5.39-4.33) ± t a/2 * sqrt((1.320201/255)+(1.032256/1169)]
= [ (1.06) ± t a/2 * 0.077848]
= [0.877758 , 1.242242]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 98% sure that the interval [0.877758 , 1.242242] 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