Question

The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to 1987. Let x₁ be a random variable that represents the time interval (in minutes) between Old Faithful eruptions for the years 1948 to 1952. Based on 9580 observations, the sample mean interval was x₁ = 61.8 minutes. Let x₂ be a random variable that represents the time interval in minutes between Old Faithful eruptions for the years 1983 to 1987. Based on 23,000 observations, the sample mean time interval was x₂ = 69.2 minutes. Historical data suggest that σ₁ = 8.49 minutes and σ₂ = 11.78 minutes. Let μ₁ be the population mean of x₁ and let μ₂ be the population mean of x₂.(a) Compute a 99% confidence interval for μ₁ – μ₂. (Use 2 decimal places.)

lower limit
upper limit

(b) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the 99% confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959.

Because the interval contains only positive numbers, we can say that the interval length between eruptions has gotten shorter.Because the interval contains both positive and negative numbers, we can not say that the interval length between eruptions has gotten longer.     We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers, we can say that the interval length between eruptions has gotten longer.

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=61.8
standard deviation , σ1 =8.49
population size(n1)=9580
y(mean)=69.2
standard deviation, σ2 =11.78
population size(n2)=23000
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((72.0801/9580)+(138.7684/23000))
= 0.1164
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.1164
= 0.2999
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (61.8-69.2) ± 0.2999 ]
= [-7.6999 , -7.1001]
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DIRECT METHOD
given that,
mean(x)=61.8
standard deviation , σ1 =8.49
number(n1)=9580
y(mean)=69.2
standard deviation, σ2 =11.78
number(n2)=23000
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 61.8-69.2) ±Z a/2 * Sqrt( 72.0801/9580+138.7684/23000)]
= [ (-7.4) ± Z a/2 * Sqrt( 0.0136) ]
= [ (-7.4) ± 2.576 * Sqrt( 0.0136) ]
= [-7.6999 , -7.1001]
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interpretations:
1. we are 99% sure that the interval [-7.6999 , -7.1001] contains the difference between
true population mean U1 - U2
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 difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.01 true mean
difference is zero
Answer:
a.
99% sure that the interval (-7.69,-7.10)
b.
yes,
interval consist of only negative numbers
We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers,
we can say that the interval length between eruptions has gotten longer.