1. The percentage of ash oil from a certain region has a normal distribution wit

Question

1. The percentage of ash oil from a certain region has a normal distribution with mean,Mu, and a standard deviation. The percentages of ash in coal in a random sample from this second region has n=20, sample mean=12.12 and sample standard deviation(s)=3.27.

A. Use the noraml probability rue to obtain an interval that contains approximately 95% of the percentages of ash for coal from this region.

B. Compute a 90% confidence interval for (standard deviation)^2

C. Compute a 90% confidence interval for the standard devation.

2. As from question 1, n=20, sample mean=12.12, and sample standard devation(s)=3.27

A. Find a 95% comfidence interval for the mean, Mu.

B. In part (a), what is meant by 95% confidence?

C. Why is the answer in part (a) in this problem different than part (a) from question 1?

Explanation / Answer

Confidence Interval
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
Mean(x)=12.12
Standard deviation( sd )=3.27
Sample Size(n)=20
Confidence Interval = [ 12.12 ± t a/2 ( 3.27/ Sqrt ( 20) ) ]
= [ 12.12 - 2.093 * (0.731) , 12.12 + 2.093 * (0.731) ]
= [ 10.59,13.65 ]

b.
Confidence Interval
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S^2 = Variance
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.1
^2 right = (1 - Confidence Level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
^2 left = 1 - ^2 right = 1 - 0.05 = 0.95
the two critical values ^2 left, ^2 right at 19 df are 30.1435 , 10.117
Variacne( S^2 )=10.6929
Sample Size(n)=20
Confidence Interval = [ 19 * 10.6929/30.1435 < ^2 < 19 * 10.6929/10.117 ]
= [ 203.1651/30.1435 < ^2 < 203.1651/10.117 ]
[ 6.7399 , 20.0816 ]

c.
confidance interval for sd=[ sqrt(6.7399) , sqrt(20.0816) ] = [ 2.596, 4.481]

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