Hypothesis Tests and Connidence Intervals for one and two samples. MEANS. Height
ID: 3311606 • Letter: H
Question
Hypothesis Tests and Connidence Intervals for one and two samples. MEANS. Heights ofta" bullius 30 or more storles in a lange city ls at least 700 feet high. A random sample of 10 balldings is selected, and the heights an feet ore shown. At _ a.cs, is there erou.gh evidence to reject the clalm? Aresearcher estimates that the "erage hel ht of buldings of · (b) Find a 95%Confidence hterval for the true height ofthese buildings. (c) Find n if you want E to be 60. 2. Retention Test Scores A random sample of non-English majors at a selected college wes used in a study to see if the student retained more from reading a 19 century novel or by watching it in DVD form. Each student was assigned one novel to read and a ditterent one to watch, and then they were glven 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? 7S 90 84 75 80 DVD 85 280 (b) Find·95% Cmdence Interval for the true difference In the means. V/m Use: df n+2-2 - +
Explanation / Answer
Q1.
PART A.
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
PART B.
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
= (213.8/60 ) ^2
= 12.7 ~ 13
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.