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

Topic 7 Exercises Kristin Barner Chapter 19 19.9 Randomly selected records of 140 convicted criminals reveal that their crimes were committed on the following days of the week: (a) Using the .01 level of significanc ...

Topic 7 Exercises Kristin Barner Chapter 19 19.9 Randomly selected records of 140 convicted criminals reveal that their crimes were committed on the following days of the week: (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. H1: H0 is false H0= Psun = Pmon = Ptues = Pwed = Pthurs = Pfri = Psat = 1/7 We reject H0 at the 0.01 level of significance = X 2 > 16.81 Fe = (expected proportion)(total sample size) frequenc y Mon Tues Wed Thurs Fri Sat Sun Total f0 17 21 22 18 23 24 15 140 fe 20 20 20 20 20 20 20 140 Null Hypothesis: X 2 = ∑ f 0−fe (¿¿ fe)2 ¿ = { (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 The null hypothesis is rejected at the 0.01 significance level because the observed X2 at 3.4 is less than the critical X2 at 16.81 It can be said that crimes are likely to be committed on any day of the week, not one more than another. (b) Specify the approximate p-value for this test result. This study source was downloaded by 100000831988016 from CourseHero.com on 05-02-2022 07:23:02 GMT -05:00 https://www.coursehero.com/file/43495891/Topic-7-Exercisesdocx/ The table indicates the value of 12.6 at the .05 level and degrees of freedom of 6. This means that the null hypothesis can be retained at the .05 level. p > .05. (c) How might this result be reported in the literature? In the literature, it may be reported that crimes are likely to occur on any day of the week. There is not one day of the week that is more likely to have crimes committed. 19.10 A number of investigators have reported a tendency for more people to die (from natural causes, such as cancer and strokes) after, rather than before, a major holiday. This post-holiday death peak has been attributed to a number of factors, including the willful postponement of death until after the holiday, as well as holiday stress and post-holiday depression. Writing in the Journal of the American Medical Association (April 11, 1990), Phillips and Smith report that among a total of 103 elderly California women of Chinese descent who died of natural causes within one week of the Harvest Moon Festival, only 33 died the week before, while 70 died the week after. (a) Using the .05 level of significance, test the null hypothesis that, in the underlying population, people are equally likely to die either the week before or the week after this holiday. Within a week before: 33 Week after: 70 N = 103 P1 = P2 = ½ E (X1)2 E (X2) = 103/2 = 51.5 E 2 = (33 – 51.5)+(70 -51.5)2 = 13.291 X = .05 critical = 3.81 TS > critical Because of the data, we can reject the null hypothesis. We can then conclude that there is significance data to show that more people die after the holiday than before the holiday. (b) Specify the approximate p-value for this test result. p = .05 p > .05 19.13 Students are classified according to religious preference (Buddhist, Jewish, Protestant, Roman Catholic, or Other) and political affiliation (Democrat, Republican, Independent, or Other). (a) Is anything suspicious about these observed frequencies? Yes, there is something suspicious about these observed frequencies. In this problem, all of the frequencies end in multiples of 10. This makes me think that the observed frequencies may not be real and may be being used to simplify the data and calculations. This study source was downloaded by 100000831988016 from CourseHero.com on 05-02-2022 07:23:02 GMT -05:00 https://www.coursehero.com/file/43495891/Topic-7-Exercisesdocx/ (b) Using the .05 level of significance, test the null hypothesis that these two variables are independent. Statistical hypothesis: H0: Political affiliation and religious preferences are independent factors H1: H0 is not true X 2 = { (30−20) 2 20 + (30−20) 2 20 + (30−20) 2 20 + (18−20)2 20 + (100−40)2 40 } = { (30−20) 2 20 + (30−20) 2 20 + (40−20)2 20 + (30−20) 2 20 + (20−20)2 20 + (20−20) 2 20 + (10−20) 2 20 } = 220 X 2 = 220 is larger than 21.03. This means that there is a relationship between the religious preferences of students and their political affiliation. (c) If appropriate, estimate the effect size. Ø 2 c = 220 500 (4−1) = 0.15 Effect size: Medium 19.14 In 1912 over 800 passengers perished after the ocean liner Titanic collided with an iceberg and sank. The following table compares the survival frequencies of cabin and steerage passengers. (a) Using the .05 level of significance, test the null hypothesis that survival rates are independent of the passengers’ accommodations (cabin or steerage). Statistical hypothesis: H0 : type of accommodations and survival rates are independent H1 : H0 is false = (c-1) (r-1) = (2-1) (2-1) = 1 fe (column total)(rowtotal) grand total fe (cabin, survived) = (579)(485) 1291 = 280815 1291 = 217.52 fe (steerage, survived) = (712)(485) 1291 = 345320 1291 = 276.48 This study source was downloaded by 100000831988016 from CourseHero.com on 05-02-2022 07:23:02 GMT -05:00 https://www.coursehero.com/file/43495891/Topic-7-Exercisesdocx/ fe (cabin, not survived) = (579)(806) 1291 = 466674 1291 = 361.48 fe (steerage, not survived) = (712)(806) 1291 = 573872 1291 = 444.52 Survived/Accommodations Cabin Steerage Total Yes f0 fe 299 217.52 186 276.48 485 No f0 fe 280 361.48 526 444.52 806 Total 579 712 1291 X 2 = (299−217.52) 2 217.52 + (186−267.48) 2 267.48 + (280−231.48) 2 361.48 + (516−444.52)2 444.52 = 6638.99 217.52 + 6638.99 267.48 + 6638.99 361.48 + 6638.99 444.52 = 30.52 + 24.82 + 18.37+ 14.94 = 88.65 The null hypothesis will be rejected because x2 of 88.65 > critical x2 of 3.84. There seems to be a correlation between the type of accommodation and the survival rate. (b) Assuming a significant c2, estimate the strength of the relationship. Ø 2 c = 88.65 1291(2−1) = 88.65 1291 = 0.07 Strength of relationship = medium (c) To more fully appreciate the importance of this relationship, calculate an odds ratio to determine how much more likely a cabin passenger is to have survived than a steerage passenger. OR = 299/280 186 /526 = 1.07 0.35 = 3.06 A cabin passenger is 3.06 times more likely to survive than a steerage passenger. 19.16 Test the null hypothesis at the .01 level of significance that the distribution of blood types for college students complies with the proportions described in the blood bank bulletin, namely, .44 for O, .41 for A, .10 for B, and .05 for AB. Now, however, assume that the results are available for a random sample of only 60 students. The results are as follows: 27 for O, 26 for A, 4 for B, and 3 for AB. Since the p value in this problem is 0.7642 is greater than .01 ( 0.7642 > .01) the test is not considered significant. Because of this, we cannot reject the null hypothesis. Chapter 20 This study source was downloaded by 100000831988016 from CourseHero.com on 0