You wish to explore the relationship between two variables, X and Y, for 17 pati

Question

You wish to explore the relationship between two variables, X and Y, for 17 patients for purposes of predicting the value of Y, given the value of X. In order (from patient 1 through patient 17), their corresponding values for X are: 9, 15, 10, 16, 10, 20, 11, 20, 15, 15, 8, 13, 18, 10, 8, 10, 16 and the matching (corresponding) values of Y: 45, 57, 45, 51, 65, 88, 44, 87, 89, 59, 66, 65, 56, 47, 66, 41, 56. Given this information, calculate the lower bound of the confidence interval of the slope, assuming an alpha of 0.01. Provide your answer with a level of accuracy of 1 decimal place.

Explanation / Answer

TRADITIONAL METHOD
given that,
mean(x)=13.1765
standard deviation , s.d1=4.0193
number(n1)=17
y(mean)=60.4118
standard deviation, s.d2 =15.476
number(n2)=17
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((16.2/17)+(239.5/17))
= 3.9
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 16 d.f is 2.9
margin of error = 2.92 * 3.9
= 11.3
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (13.1765-60.4118) ± 11.3 ]
= [-58.6 , -35.9]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=13.1765
standard deviation , s.d1=4.0193
sample size, n1=17
y(mean)=60.4118
standard deviation, s.d2 =15.476
sample size,n2 =17
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 = [( 13.1765-60.4118) ± t a/2 * sqrt((16.2/17)+(239.5/17)]
= [ (-47.2) ± t a/2 * 3.9]
= [-58.6 , -35.9]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [-58.6 , -35.9] 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

lower bound interval = -58.6

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