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5 votes
A data set lists earthquake depths. The summary statistics are

nequals=400400​,
x overbarxequals=6.866.86
​km,
sequals=4.374.37
km. Use a
0.010.01
significance level to test the claim of a seismologist that these earthquakes are from a population with a mean equal to
6.006.00.
Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.
What are the null and alternative​ hypotheses?


A.
Upper H 0H0​:
muμequals=5.005.00
km
Upper H 1H1​:
muμnot equals≠5.005.00
km

B.
Upper H 0H0​:
muμnot equals≠5.005.00
km
Upper H 1H1​:
muμequals=5.005.00
km

C.
Upper H 0H0​:
muμequals=5.005.00
km
Upper H 1H1​:
muμgreater than>5.005.00
km

D.
Upper H 0H0​:
muμequals=5.005.00
km
Upper H 1H1​:
muμless than<5.005.00
km
Determine the test statistic.


​(Round to two decimal places as​ needed.)
Determine the​ P-value.


​(Round to three decimal places as​ needed.)
State the final conclusion that addresses the original claim.

Fail to reject

Upper H 0H0.
There is


evidence to conclude that the original claim that the mean of the population of earthquake depths is
5.005.00
km

1 Answer

2 votes

Answer:

Explanation:

The summary of the given statistics data include:

sample size n = 400

sample mean
\overline x = 6.86

standard deviation = 4.37

Level of significance ∝ = 0.01

Population Mean
\mu = 6.00

Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

To start with the hypothesis;

The null and the alternative hypothesis can be computed as :


H_o: \mu = 6.00 \\ \\ H_1 : \mu \\eq 6.00

The test statistics for this two tailed test can be computed as:


z= \frac{\overline x - \mu}{\frac{\sigma}{\sqrt {n}}}


z= \frac{6.86 - 6.00}{\frac{4.37}{\sqrt {400}}}


z= (0.86)/((4.37)/(20))

z = 3.936

degree of freedom = n - 1

degree of freedom = 400 - 1

degree of freedom = 399

At the level of significance ∝ = 0.01

P -value = 2 × (z < 3.936) since it is a two tailed test

P -value = 2 × ( 1 - P(z ≤ 3.936)

P -value = 2 × ( 1 -0.9999)

P -value = 2 × ( 0.0001)

P -value = 0.0002

Since the P-value is less than level of significance , we reject
H_o at level of significance 0.01

Conclusion: There is sufficient evidence to conclude that the original claim that the mean of the population of earthquake depths is 5.00 km.

User Zimbabwe Elephant
by
8.3k points
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