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

Respuesta :

Answer:

Step-by-step explanation:

The summary of the given statistics data include:

sample size n = 400

sample mean [tex]\overline x[/tex] = 6.86

standard deviation = 4.37

Level of significance ∝ = 0.01

Population Mean [tex]\mu[/tex] = 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 :

[tex]H_o: \mu = 6.00 \\ \\ H_1 : \mu \neq 6.00[/tex]

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

[tex]z= \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt {n}}}[/tex]

[tex]z= \dfrac{6.86 - 6.00}{\dfrac{4.37}{\sqrt {400}}}[/tex]

[tex]z= \dfrac{0.86}{\dfrac{4.37}{20}}[/tex]

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 [tex]H_o[/tex] 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.