PSY 520 Topic 7 Exercise:Chapter 19 and 20-Latest Updated 2022 Already Passed

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 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 24 15 140 f e 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. 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 H0 : Pheads=Ptails= 1 248 Decision rule: Reject H0 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 f o 30 20 50 f e 25 25 50 Null hypothesis: x 2=∑ (f o−f e )2 f e ¿ { (30−25) 2 25 + (20−25) 2 25 } = 25 25 + 25 25 = 1+1 = 2 Retain because x 2 of 2 is less than critical x 2 of 3.84. b.) Specify the approximate p-value for this test result. p-value is the smallest level of significance that would lead to the rejection of H0 . Acceptance of it would lead to a greater p-value than significance level. At the 5% level, we retain H0 implies that the p-value is greater than a = 0.05. 19.13 a.) Using the .05 level of significance test the null hypothesis that survival rates are independent of the passengers’ accommodations (cabin or steerage). H0 : type of accommodations and survival rates are independent H1 : H0 isfalse Decision Rule: Reject the null hypothesis at 0.05 level of significance, if x 2 ≥3.84 , given that the degrees of freedom. =(c-1)(r-1) =(2-1)(2-1) =1 Calculations: f e= ( columntotal)(rowtotal) grandtotal f e ( cabin,survived)= (579)(485) 1291 = 1291 =217.52 f e (steerage ,survived)= (712) (485) 1291 = 1291 =276.48