Question

Question 1

1. It is known that the weights of Persian cats form a normal distribution. A random sample of 20 Persian cats was collected and the sample mean and the standard deviation were 4.5kg and 1.2kg, respectively.

2. It is known that the weights of Persian cats form a normal distribution. A random sample of 20 Persian cats was collected and the sample mean and the standard deviation were 4.5kg and 1.2kg, respectively.

3. 1. Construct an interval where the population mean will lie with 95% confidence.
      
   2. Construct an interval that will contain about 95% of the data.
      
   3. If there is a cat with its weight of 1.8kg, can we consider this observation as an outlier?

Answer 1

1.

TRADITIONAL METHOD
given that,
sample mean, x =4.5
standard deviation, s =1.2
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.2/ sqrt ( 20) )
= 0.27
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 0.27
= 0.56
III.
CI = x ± margin of error
confidence interval = [ 4.5 ± 0.56 ]
= [ 3.94 , 5.06 ]
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DIRECT METHOD
given that,
sample mean, x =4.5
standard deviation, s =1.2
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/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 = [ 4.5 ± t a/2 ( 1.2/ Sqrt ( 20) ]
= [ 4.5-(2.093 * 0.27) , 4.5+(2.093 * 0.27) ]
= [ 3.94 , 5.06 ]
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interpretations:
1) we are 95% sure that the interval [ 3.94 , 5.06 ] 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
Answer:
a.
95% sure that the interval [ 3.94 , 5.06 ] contains the true population mean
interval where the population mean will lie with 95% confidence.
b.
interval that will contain about 95% of the data.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 4.5
standard Deviation ( sd )= 1.2
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [4.5 ± 2 * 1.2]
= [ 4.5 - 2 * 1.2 , 4.5 + 2* 1.2]
= [ 2.1 , 6.9 ]
c.
If there is a cat with its weight of 1.8kg,
yes,
this observation as an outlier because 1.8kg is not in the range of interval.