8.3 A researcher recently examined the studying habits of college students. It i

Question

8.3 A researcher recently examined the studying habits of college students. It is known that the number of hours per week spent studying by Communications majors follows a N(1,1) distribution. Also, it is known that the number of hours per week spent studying by English majors follows a N(2,2) distribution. A random sample of 16 Communications majors and 26 English majors found: Communications: English: n1=16 n2=26 x1=5.5 x2=7.3 s1=2.0 s2=2.4 Assume the population standard deviations are equal, i.e. 1-2 (a) Suppose we wish to test Ho : 1 = 2 vs. Ha : 12 at the = 0.01 significance level. Approximate the p-value for this test using the t-table. (b) (3 pts) Based upon your answer in 8.3(a), is there a significant difference in the population mean study times? Why?

Explanation / Answer

8.3)a.
Given that,
mean(x)=5.5
standard deviation , s.d1=2
number(n1)=16
y(mean)=7.3
standard deviation, s.d2 =2.4
number(n2)=26
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.947
since our test is two-tailed
reject Ho, if to < -2.947 OR if to > 2.947
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =5.5-7.3/sqrt((4/16)+(5.76/26))
to =-2.6213
| to | =2.6213
critical value
the value of |t | with min (n1-1, n2-1) i.e 15 d.f is 2.947
we got |to| = 2.62128 & | t | = 2.947
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.6213 ) = 0.019
hence value of p0.01 < 0.019,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.6213
critical value: -2.947 , 2.947
decision: do not reject Ho
p-value: 0.019
b.
we do not have enough evidence to support the claim and there is significant difference in the population mean study times

d.
TRADITIONAL METHOD
given that,
mean(x)=5.5
standard deviation , s.d1=2
number(n1)=16
y(mean)=7.3
standard deviation, s.d2 =2.4
number(n2)=26
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((4/16)+(5.76/26))
= 0.687
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 15 d.f is 2.947
margin of error = 2.947 * 0.687
= 2.024
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.5-7.3) ± 2.024 ]
= [-3.824 , 0.224]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=5.5
standard deviation , s.d1=2
sample size, n1=16
y(mean)=7.3
standard deviation, s.d2 =2.4
sample size,n2 =26
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 = [( 5.5-7.3) ± t a/2 * sqrt((4/16)+(5.76/26)]
= [ (-1.8) ± t a/2 * 0.687]
= [-3.824 , 0.224]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-3.824 , 0.224] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion

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