19.9
a.) Using the .01 level of significance, test the null hypothesis that in the
underlying population, crimes are equally likely to be committed on any day
of the week.
Response:
H0=Psun=Pmon=Ptues=Pwed=Pthurs=Pfri=Psat=
1
7
H1
: H0
isfalse
Decision Rule: Reject H0 at the 0.01 level o...
19.9
a.) Using the .01 level of significance, test the null hypothesis that in the
underlying population, crimes are equally likely to be committed on any day
of the week.
Response:
1
H 0=P sun=Pmon=P tues=P wed =Pthurs=P fri=P sat =
7
H 1 : H 0 is false
Decision Rule: Reject H0 at the 0.01 level of significance is
2
x ≥ 16.81
Calculations:
total sample ¿ ¿
f e =( expected proportion ) ¿
Frequen Mon Tues Wed Thurs Fri Sat Sun Total
cy
fo 17 21 22 18 23 24 15 140
fe 20 20 20 20 20 20 20 140
= 3.4
2
Return the null hypothesis at 0.01 level because the observed x of 3.4 is
2
smaller than the critical x of 16.81. Crimes are likely to be committed on
any day of the week.
b.) Specify the approximate p -value for this test result.
, At 1% level significance, we return the null hypothesis H0 indicates the p-
2
value is greater than 0.01, that is, p>0.01. In the x table values we
observe the value 12.6 at 0.05 level and degrees of freedom = 6. We can
retain the null hypothesis at 5% level too. Therefore p>0.05.
c.) How might this result be reported in literature?
There is evidence that the crimes are equally likely to take place on any day
of the week [ x ( 6, n=200 )=3.4, p> 0.05 ] . We are unable to calculate ∅c , since
2 2
non-significant x 2 at 0.01 level. The parenthetical statement indicates that
2
a x based on 6 degrees of freedom and a sample size of 140 was found to
equal 3.4. The test result has an approximate p-value greater than 0.05,
because the null hypothesis was retained.
19.10
a.) Test the null hypothesis that this coin is unbiased, that is, that heads and tails are
equally likely to appear in the long run.
Null hypothesis
1
H 0 : P heads=P tails=
2 48
Decision rule: Reject H 0 at the 0.05 level of significance; if x2 ≥3.84 ,
given that the degrees of freedom equals to
=c-1
=2-1
=1
Calculations:
Frequency Heads Tails Total
fo 30 20 50
fe 25 25 50
Null hypothesis:
( f −f )2
x 2= ∑ o e
fe
( 30−25 )2 (20−25 )2
¿{ + }
25 25
25 25
= 25 + 25
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller BrilliantScores. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $12.49. You're not tied to anything after your purchase.
