AN thesis Tests and Confidence Intervals for one and two samples. ME Hypo of t.
ID: 3311621 • Letter: A
Question
AN thesis Tests and Confidence Intervals for one and two samples. ME Hypo of t. Heights of tall buildings A researcher estimates that the average height of buildings 30 or more stories in a lange city is at least 700 feet high. A random sample of 10 buildings is selected, andthe heights in feet are shown. At -0.05, is there enough evidence to rejact the claim? as-1520-1535-1635-1616 (b) Find a 95% Confidence Interval for the true height of these buildings. (c) Flnd n if you want E to be 6o. 2. Retention Test Scores A random sample of non-English majors at a selected coll used in a study to see if the student retained more from reading a 19th by watch one to watch, and then they were elven a hundred point quiz on each novel. The test results are shown. At 0.05, can it be concluded that the book scores are higher than the DVD scores? century novel or ing it in DVD form. Each student was assigned one novel to read and a different Book90 80 90 DVD 85 7280 75 90 75 70 80 (b) Find a 95% Confidence Interval for the true difference In the means. 9/2S /n 72 e: df=n1+n2-2 1-9 o 2 mi 2Explanation / Answer
1.
a.
Given that,
population mean(u)=700
sample mean, x =606.5
standard deviation, s =109.0792
number (n)=10
null, Ho: >=700
alternate, H1: <700
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.8331
since our test is left-tailed
reject Ho, if to < -1.8331
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =606.5-700/(109.0792/sqrt(10))
to =-2.711
| to | =2.711
critical value
the value of |t | with n-1 = 9 d.f is 1.8331
we got |to| =2.711 & | t | =1.8331
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.7106 ) = 0.01199
hence value of p0.05 > 0.01199,here we reject Ho
ANSWERS
---------------
null, Ho: >=700
alternate, H1: <700
test statistic: -2.711
critical value: -1.8331
decision: reject Ho
p-value: 0.01199
b.
TRADITIONAL METHOD
given that,
sample mean, x =606.5
standard deviation, s =109.0792
sample size, n =10
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 109.0792/ sqrt ( 10) )
= 34.494
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262
margin of error = 2.262 * 34.494
= 78.025
III.
CI = x ± margin of error
confidence interval = [ 606.5 ± 78.025 ]
= [ 528.475 , 684.525 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =606.5
standard deviation, s =109.0792
sample size, n =10
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 2.262
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 = [ 606.5 ± t a/2 ( 109.0792/ Sqrt ( 10) ]
= [ 606.5-(2.262 * 34.494) , 606.5+(2.262 * 34.494) ]
= [ 528.475 , 684.525 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 528.475 , 684.525 ] 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.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 109.0792
ME =60
n = ( 1.96*109.0792/60) ^2
= (213.795/60 ) ^2
= 12.697 ~ 13
2.
a.
Given that,
mean(x)=84.1429
standard deviation , s.d1=6.0671
number(n1)=7
y(mean)=77.4286
standard deviation, s.d2 =5.287
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.94
since our test is right-tailed
reject Ho, if to > 1.94
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =84.1429-77.4286/sqrt((36.8097/7)+(27.95237/7))
to =2.21
| to | =2.21
critical value
the value of |t | with min (n1-1, n2-1) i.e 6 d.f is 1.94
we got |to| = 2.20744 & | t | = 1.94
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.2074 ) = 0.03469
hence value of p0.05 > 0.03469,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.21
critical value: 1.94
decision: reject Ho
p-value: 0.03469
we have enough evidence to support that books scores are higher than DVD scores
b.
TRADITIONAL METHOD
given that,
mean(x)=84.1429
standard deviation , s.d1=6.0671
number(n1)=7
y(mean)=77.4286
standard deviation, s.d2 =5.287
number(n2)=7
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((36.81/7)+(27.952/7))
= 3.042
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 6 d.f is 2.447
margin of error = 2.447 * 3.042
= 7.443
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (84.1429-77.4286) ± 7.443 ]
= [-0.729 , 14.157]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=84.1429
standard deviation , s.d1=6.0671
sample size, n1=7
y(mean)=77.4286
standard deviation, s.d2 =5.287
sample size,n2 =7
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 = [( 84.1429-77.4286) ± t a/2 * sqrt((36.81/7)+(27.952/7)]
= [ (6.714) ± t a/2 * 3.042]
= [-0.729 , 14.157]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-0.729 , 14.157] 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
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