1. Find a 95% confidence interval for the mean for both the explanatory (Foot le
ID: 3314572 • Letter: 1
Question
1. Find a 95% confidence interval for the mean for both the explanatory (Foot length) and response variables (Height).
2. For the variable Foot length, and using a level of significance = 0.05: Test the claim that the mean > 25 cm using the p-value method.
4. Calculate the correlation coefficient r.
5. Determine if there is significant linear correlation at the alpha = 0.05 significance level.
6. Predict the height of a person with a Foot length of 26cm.
Foot length Height 27.8 180.3 25.7 175.3 26.7 184.8 25.9 177.8 26.4 182.3 29.2 185.4 26.8 180.3 28.1 175.3 25.4 177.8 27.9 185.4 27.5 190.5 28.8 195 26.7 175.3 26.7 180.3 25.1 172.7 28.7 182.9 29.2 189.2 27.9 185.4 28.6 193.7 23.2 165.1 24.3 166.4 26 177.8 23.8 167.6 25.1 168.3 25.4 165.7 21.9 165.1 26.2 165.1 23.8 165.1 22.2 152.4 24.6 162.6 24.6 179.1 23.7 175.9 25.6 166.4 24.1 167.6 23.8 162.61. Find a 95% confidence interval for the mean for both the explanatory (Foot length) and response variables (Height).
2. For the variable Foot length, and using a level of significance = 0.05: Test the claim that the mean > 25 cm using the p-value method.
4. Calculate the correlation coefficient r.
5. Determine if there is significant linear correlation at the alpha = 0.05 significance level.
6. Predict the height of a person with a Foot length of 26cm.
Explanation / Answer
Q1.
TRADITIONAL METHOD
given that,
mean(x)=25.926
standard deviation , s.d1=1.982
number(n1)=35
y(mean)=175.5
standard deviation, s.d2 =10.072
number(n2)=35
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((3.928/35)+(101.445/35))
= 1.735
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 34 d.f is 2.032
margin of error = 2.032 * 1.735
= 3.526
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (25.926-175.5) ± 3.526 ]
= [-153.1 , -146.048]
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DIRECT METHOD
given that,
mean(x)=25.926
standard deviation , s.d1=1.982
sample size, n1=35
y(mean)=175.5
standard deviation, s.d2 =10.072
sample size,n2 =35
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 = [( 25.926-175.5) ± t a/2 * sqrt((3.928/35)+(101.445/35)]
= [ (-149.574) ± t a/2 * 1.735]
= [-153.1 , -146.048]
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interpretations:
1. we are 95% sure that the interval [-153.1 , -146.048] 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
Q2.
Given that,
population mean(u)=25
sample mean, x =25.926
standard deviation, s =1.982
number (n)=35
null, Ho: =25
alternate, H1: >25
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.691
since our test is right-tailed
reject Ho, if to > 1.691
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =25.926-25/(1.982/sqrt(35))
to =2.764
| to | =2.764
critical value
the value of |t | with n-1 = 34 d.f is 1.691
we got |to| =2.764 & | t | =1.691
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 2.764 ) = 0.00458
hence value of p0.05 > 0.00458,here we reject Ho
ANSWERS
---------------
null, Ho: =25
alternate, H1: >25
test statistic: 2.764
critical value: 1.691
decision: reject Ho
p-value: 0.00458
have evidence that the mean > 25 cm using the p-value method
Q3.
Correlation: X, Y
Pearson correlation of X and Y = 0.837
Q4.
Given that,
value of r =0.8365
number (n)=35
null, Ho: =0
alternate, H1: !=0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.035
since our test is two-tailed
reject Ho, if to < -2.035 OR if to > 2.035
we use test statistic (t) = r / sqrt(1-r^2/(n-2))
to=0.8365/(sqrt( ( 1-0.8365^2 )/(35-2) )
to =8.77
|to | =8.77
critical value
the value of |t | at los 0.05% is 2.035
we got |to| =8.77 & | t | =2.035
make decision
hence value of | to | > | t | and here we reject Ho
ANSWERS
---------------
null, Ho: =0
alternate, H1: !=0
test statistic: 8.77
critical value: -2.035 , 2.035
decision: reject Ho
significant correlation b/w them
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