Question

18.

The accompanying data table lists the magnitudes of 50 earthquakes measured on the Richter scale. Test the claim that the population of earthquakes has a mean magnitude greater than 1.00. Use a 0.05 significance level. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, and conclusion for the test. Assume this is a simple random sample

Magnitude of Earthquake
0.690	0.740	0.640	0.390	0.700	2.200	1.980	0.640	1.220	0.200
1.640	1.320	2.950	0.900	1.760	1.010	1.260	0.000	0.650	1.460
1.620	1.830	0.990	1.560	0.410	1.280	0.830	1.340	0.540	1.250
0.920	1.000	0.790	0.790	1.440	1.000	2.240	2.500	1.790	1.250
1.490	0.840	1.420	1.000	1.250	1.420	1.350	0.930	0.400	1.390

What are the​ hypotheses?

A.

H0​:

μ≠1.00

in magnitude

H1​:

μ=1.00

in magnitude

B.

H0​:

μ=1.00

in magnitude

H1​:

μ>1.00

in magnitude

C.

H0​:

μ=1.00

in magnitude

H1​:

μ≠1.00

in magnitude

D.

H0​:

μ=1.00

in magnitude

H1​:

μ<1.00

in magnitude

__________________

Identify the test statistic.

t= (Round to two decimal places as​ needed.)

Identify the​ P-value.

The​ P-value is (Round to three decimal places as​ needed.)

_______________________________

Choose the correct answer below.

A.

Reject

H0.

There is

sufficient

evidence to conclude that the population of earthquakes has a mean magnitude greater than

1.00.

B.

Fail to reject

H0.

There is

sufficient

evidence to conclude that the population of earthquakes has a mean magnitude greater than

1.00.

C.

Reject

H0.

There is

insufficient

evidence to conclude that the population of earthquakes has a mean magnitude greater than

1.00.

D.

Fail to reject

H0.

There is

insufficient

evidence to conclude that the population of earthquakes has a mean magnitude greater than

1.00.

Answer 1

for hypothesis: option B is correct

null hypothesis: HO: μ		=	1
Alternate Hypothesis: Ha: μ		>	1

sample mean 'x̄=		1.184
sample size    n=		50
std deviation s=		0.5872
std error ='sx=s/√n=0.587214420386799/√50=		0.0830
t statistic ='(x̄-μ)/sx=(1.1842-1)/0.083=		2.22
p value      =		0.016		from excel: tdist(2.218,49,1)

since p value <0.05

option A is correct :

Reject Ho ; there is sufficient evidence ........