please answer all parts. It is a little out of order sorry about that. i Safari

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please answer all parts. It is a little out of order sorry about that.

i Safari File Edit View History Bookmarks Window Help 496 D Thu 5:39 PM a E webassign.net Homework 7.1 Chegg Study | Guided Solutions and Study Help | Chegg.com degreee of freedom calculator - Google Search Specify the alternate hypothesis. O Hai = 16 Ha: > 16 Carry out the test. (Round your answer for tto three decimal places.) Give the degrees of freedom Give the P-value. (Round your answer to four decimal places.) What do you conclude? We have sufficient evidence to conclude that the mean weight change is not 16 pounds. We do not have sufficient evidence to conclude that the mean weight change is not 16 pounds. (f) Write a short paragraph explaining your results. This answer has not been graded yet. eBook 2 708 2

Explanation / Answer

a.
yes,it is appropriate to analyze the data using methods based on normal distribution.there are no outer lines and no particular skewness
b.
sample mean X=42.12kg
sample standard deviation s= 4.3365kg
c.
I.stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.3365/ sqrt ( 20) )
= 0.97
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 = 19 d.f is 2.093
margin of error = 2.093 * 0.97
= 2.03
DIRECT METHOD
given that,
sample mean, x =42.12
standard deviation, s =4.3365
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
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 = [ 42.12 ± t a/2 ( 4.3365/ Sqrt ( 20) ]
= [ 42.12-(2.093 * 0.97) , 42.12+(2.093 * 0.97) ]
= [ 40.09 , 44.15 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 40.09 , 44.15 ] 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
d. coverting into pounds
sample mean X=42.12*2.2 = 92.664 pounds
sample standard deviation s= 4.3365*2.2 = 9.5403 pounds
TRADITIONAL METHOD
given that,
sample mean, x =92.664
standard deviation, s =9.5403
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 9.5403/ sqrt ( 20) )
= 2.133
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 = 19 d.f is 2.093
margin of error = 2.093 * 2.133
= 4.465
III.
CI = x ± margin of error
confidence interval = [ 92.664 ± 4.465 ]
= [ 88.199 , 97.129 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =92.664
standard deviation, s =9.5403
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
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 = [ 92.664 ± t a/2 ( 9.5403/ Sqrt ( 20) ]
= [ 92.664-(2.093 * 2.133) , 92.664+(2.093 * 2.133) ]
= [ 88.199 , 97.129 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 88.199 , 97.129 ] 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
e.
Given that,
population mean(u)=16
sample mean, x =92.664
standard deviation, s =9.5403
number (n)=20
null, Ho: =16
alternate, H1: >16
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.7291
since our test is right-tailed
reject Ho, if to > 1.7291
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =92.664-16/(9.5403/sqrt(20))
to =35.937
| to | =35.937
critical value
the value of |t | with n-1 = 19 d.f is 1.7291
we got |to| =35.937 & | t | =1.7291
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 35.9372 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =16
alternate, H1: >16
test statistic: 35.937
critical value: 1.7291
decision: reject Ho
p-value: 0
we have sufficient evidence to conclude that mean weight change is not 16 pounds

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