Statistics > QUESTIONS & ANSWERS > PSY 520 Topic 7 Exercise:Chapter 19 and 20-Latest Update (All)

PSY 520 Topic 7 Exercise:Chapter 19 and 20-Latest Update

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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 of significance is x 2 ≥16.81 Calculations: totalsample¿ ¿ f e=( expected proportion) ¿ Frequen cy Mon Tues Wed Thurs Fri Sat Sun Total f o 17 21 22 18 23 24 15 140 f e 20 20 20 20 20 20 20 140 Null hypothesis: x 2=∑ (f o−f e )2 f e ¿ { (17−20) 2 20 + (21−20) 2 20 + (22−20) 2 20 + (18−20) 2 20 + (23−20) 2 20 + (24−20) 2 20 + (15−20) 2 20 } = 9 20 + 1 20 + 4 20 + 4 20 + 9 20 + 16 20 + 25 20 = 68 20 = 3.4 Return the null hypothesis at 0.01 level because the observed x 2 of 3.4 is smaller than the critical x 2 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 pvalue is greater than 0.01, that is, p>0.01. In the x 2 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 2 (6,n=200)=3.4, p>0.05] . We are unable to calculate ∅c 2 , since non-significant x 2 at 0.01 level. The parenthetical statement indicates that a x 2 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. [Show More]

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