Question

ALL PARTS PLEASE AND EXPLANATION IF POSSIBLE

3. A Department of Justice report stated that convicted burglars spent an average of 18.7 months in jail for their crime. A researcher feels that this average should be higher so she randomly samples 40 such cases from court files and finds that = 21.3 months and s = 7.7 months. Test the researchers claim that the average jail time is greater than 18.7 months at the 0.05 significance level.

Null Hypothesis: _________________________________ (write in symbolic notation)

Alternative Hypothesis: ______________________________ (write in symbolic notation)

Test statistic:

p-value:

Choose one: Reject Null or Fail to Reject Null Hypothesis. Give evidence to support this decision.

Final Conclusion (interpretation):

4. The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4 parts per million (ppm). A consumer health group measured the level of the chemical in a random sample of tomatoes obtained from one producer. The levels, in ppm, are shown below.

0.31 0.47 0.19 0.72 0.56

0.91 0.29 0.83 0.49 0.28

0.31 0.46 0.25 0.34 0.17

0.58 0.19 0.26 0.47 0.81

Do the data provide sufficient evidence to support the claim that the mean level of the chemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm? Use a 0.05 significance level to test the claim that these sample levels come from a population with a mean greater than 0.4 ppm.

Initial Claim:

Null Hypothesis:

Alternative Hypothesis:

Test statistic:

p-value:

Choose one: Reject Null or Fail to Reject Null Hypothesis. Give evidence to support this decision. You may use a graph as your evidence.

Final Conclusion:

Answer 1

3)
H0 : mu = 18.7
Ha: mu > 18.7

test statistics:

z= (x -mena)/(s/sqrt(n))
= ( 21.3 - 18.7)/(77/sqrt(40))
= 0.2136

p value= .4154

p value is less than 0.05
Fail to Reject Null Hypothesis
There is not sufficient evidence to support the claim that the average jail time is greater than 18.7 months

4)

H0 ; mu = 0.4
Ha: mu > 0.4

tets statistics;
t = (x -mean)/(s/sqrt(n))
= ( 0.445 - 0.4)/(0.2277/sqrt(20))
= 0.8838

p value = .1939

do not reejct H0

There is not sufficient evidence to support the claim that these sample levels come from a population with a mean greater than 0.4 ppm.