Question

1b The Earth is structured in layers: crust, mantle, and core. A recent study was conducted to estimate the mean depth of the upper mantle in a specific farming region in California. Twenty-six, n = 26 sample sites were selected at random from a normally distributed population of approximately N = 1598 sites, and the depth of the upper mantle was measured using changes in seismic velocity and density. The sample mean was 127.5 km and the sample standard deviation was 21.3 km. Suppose the depth of the upper mantle is normally distributed. Find a 90% confidence interval for the true mean depth of the upper mantle in this farming region.

1. Determine the prerequisites have been met (3 pts):

• Random sample:
• n ≤ 0.05 of N:
• the population is normally distributed or n ≤ 30:

2. Construct the confidence interval (4 pts):

3. Interpret the confidence interval, in a statement (1 pt):

1c According to the U.S. Fire Administration, approximately N = 25,000 fires are caused by fire-works each year in the United States. Despite numerous public warnings against the use of fireworks, the home property damage due to these fires is enormous. In a random sample of n = 25 fires due to fireworks, the resulting mean property damage (in dollars) was 860.75 with a standard deviation of 350.50. Assume the underlying distribution of property damage due to these fires is normal. Find a 99% confidence interval for the true mean property damage due to a fire caused by fireworks.

1. Determine the prerequisites have been met (3 pts):

• Random sample:
• n ≤ 0.05 of N:
• the population is normally distributed or n ≤ 30:

2. Construct the confidence interval (4 pts):

3. Interpret the confidence interval, in a statement (1 pt):

Answer 1

(1b):

(a)

the prerequisites have been met:

Random sample: Yes, since stated in the description

n = Sample Size = 26 0.05 X N = 0.05 X 1598 = 79.9

the population is normally distributed , since stated in the description

(b)

SE = s/

= 21.3/

= 4.1773

ndf = n - 1 = 26 - 1 = 25

= 0.10

From Table, critical values of t = 1.7081

Confidence Interval:

127.5 (1.7081 X 4.1773)

= 127.5 7.1352

= (120.3648 ,134.6352)

Confidence Interval:

120.3648 < <   134.6352

(c)

The 90% Confidence Interval (120.3648 ,134.6352) is a range of values that we can be 90% certain contains the true mean of the population.

(1c)

(a)

the prerequisites have been met:

Random sample: Yes, since stated in the description

n = Sample Size = 25 0.05 X N = 0.05 X 25,000 = 1250

the population is normally distributed , since stated in the description

(b)

SE = s/

= 350.50/

= 70.10

ndf = n - 1 = 25 - 1 = 24

= 0.01

From Table, critical values of t = 2.7969

Confidence Interval:

860.75 (2.7969 X 70.10)

= 860.75 196.0627

= (664.69 ,1056.81)

Confidence Interval:

664.69 < <   1056.81

(c)

The 99% Confidence Interval (664.69 ,1056.81) is a range of values that we can be 99% certain contains the true mean of the population.