8. A Secchi disk is a weighted disk that is painted black and white with a rope

Question

8. A Secchi disk is a weighted disk that is painted black and white with a rope attached to it. It is used to inches at which it is no longer visible. A biologist is interested in determining the water clarity in Lake Percy Quin the clarity of the water. This is done by lowering the ball into the water and recording the depth in n is improving. He takes the measurements at the same locations on the same dates during the course of a year and repeats the measurements five years later Initial depth 38 58 65 74 56 36 56 52 Depth 5 years later 52 60 72 72 54 48 Using = 0.05, test whether the clarity of the lake is improving. Hypotheses: Ho: 58 60 Ha: Test Statistic: Graph: Conclusion: Construct a 95% C.1, for the difference:

Explanation / Answer

8.

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -5.13
We have d = -5.13
pooled variance = calculate value of Sd= S^2 = sqrt [ 469-(-41^2/8 ] / 7 = 6.08
to = d/ (S/n) = -2.39
critical Value
the value of |t | with n-1 = 7 d.f is 2.365
we got |t o| = 2.39 & |t | =2.365
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.386 ) = 0.0485
hence value of p0.05 > 0.0485,here we reject Ho
ANSWERS
---------------
a.
null, H0: Ud = 0
alternate, H1: Ud < 0
b.
test statistic: -2.39
critical value: reject Ho, if to < -2.365 OR if to > 2.365
c.
decision: Reject Ho
p-value: 0.0485

d.
TRADITIONAL METHOD
given that,
mean(x)=54.375
standard deviation , s.d1=12.6935
number(n1)=8
y(mean)=59.5
standard deviation, s.d2 =8.7342
number(n2)=8
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((161.12/8)+(76.29/8))
= 5.45
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 7 d.f is 2.36
margin of error = 2.365 * 5.45
= 12.88
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (54.375-59.5) ± 12.88 ]
= [-18.01 , 7.76]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=54.375
standard deviation , s.d1=12.6935
sample size, n1=8
y(mean)=59.5
standard deviation, s.d2 =8.7342
sample size,n2 =8
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 = [( 54.375-59.5) ± t a/2 * sqrt((161.12/8)+(76.29/8)]
= [ (-5.13) ± t a/2 * 5.45]
= [-18.01 , 7.76]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-18.01 , 7.76] 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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