Question

Assume that a simple random sample has been selected and test the given claim. Listed below are the lead concentrations in mu g divided by gμg/g measured in different traditional medicines. Use a 0.10 significance level to test the claim that the mean lead concentration for all such medicines is less than 16 mu g divided by gμg/g.

3.0 6.0 6.5 5.5 20.5 8.0 11.5 20.5 12.5 17.5

  Identify the null and alternative hypotheses for this test.

A.

Upper H 0H0​:

muμless than<1616

mu g divided by gμg/g

Upper H 1H1​:

muμequals=1616

mu g divided by gμg/g

B.

Upper H 0H0​:

muμequals=1616

mu g divided by gμg/g

Upper H 1H1​:

muμgreater than>1616

mu g divided by gμg/g

C.

Upper H 0H0​:

muμequals=1616

mu g divided by gμg/g

Upper H 1H1​:

muμless than<1616

mu g divided by gμg/g

D.

Upper H 0H0​:

muμequals=1616

mu g divided by gμg/g

Upper H 1H1​:

muμnot equals≠1616

mu g divided by gμg/g

Identify the value of the test statistic.

nothing

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

Identify the​ P-value.

nothing

​(Round to three decimal places as​ needed.)

State the conclusion about the null​ hypothesis, as well as the final conclusion that addresses the original claim.

A.

Fail to rejectFail to reject

Upper H 0H0.

There

is notis not

sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than

1616

mu g divided by gμg/g.

B.

RejectReject

Upper H 0H0.

There

is notis not

sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than

1616

mu g divided by gμg/g.

C.

RejectReject

Upper H 0H0.

There

isis

sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than

1616

mu g divided by gμg/g.

D.

Fail to rejectFail to reject

Upper H 0H0.

There

isis

sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than

1616

mu g divided by gμg/g.

Answer 1

From the given sample data : n=10 , ,

The sample mean is ,

The null and alternative hypothesis is ,

The value of the test statistic is ,

The p-value is ,

p-value= ; The Excel function is , =TDIST(2.38,9,1)

Decision : Here , p-value = 0.021<=0.10

Therefore , reject Ho

Conclusion : There is sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than 16 mu g divided by .