Grand Canyon University PSY 520 Topic 7 Exercises 100% Pass

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TOPIC 7 EXERCISES Topic 7 Exercises Angi Ware Grand Canyon University PSY 520 July 27, 2017 TOPIC 7 EXERCISES 19.9 Randomly selected records of 140 convicted criminals reveal that their crimes w... ere 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. Statistical Hypothesis: H0 = Psun = Pmon = Ptues = Pwed = Pthurs = Pfri = Psat = 1 7 H1 : H0 is false Decision Rule: Reject H0 at the 0.01 level of significance = x2 > 16.81 fe = (expected proportion) (total sample size) frequency 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 returned at the 0.01 level because the observed x2 of 3.4 is < the critical x2 of 16.81 The result is that crimes are likely to be committed on any day of the week. TOPIC 7 EXERCISES (b) Specify the approximate p -value for this test result. The null hypothesis H0 is returned at the 1% level of significance, which indicates that p > 0.01. The x2 table values indicate the value 12.6 at the 0.05 level and degrees of freedom =6. The null hypothesis can be retained at the 5% level, so p>0.05. (c) How might this result be reported in the literature? The evidence states that the crimes would equally likely occur on any day of the week. [x2 (6, n=200) = 3.4, p>0.05]. This indicates that x2 based on 6 degrees of freedom with a sample size of 140 = 3.4.The test result has an approximate p value >0.05 because the null hypothesis was retained. 19.10 While playing a coin-tossing game in which you are to guess whether heads or tails will appear, you observe 30 heads in a string of 50 coin tosses. (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 = 1 2 48 Decision rule: Reject H0 at the 0.05 level of significance. If x2>3.84 given the degrees of freedom to be: c-1=2-1=1 Frequency Heads Tails Total f0 30 20 50 fe 25 25 50 Null hypothesis: X 2 = ∑ ( f 0−fe) 2 fe = { (30−25) 2 25 + (20−25) 2 25 } TOPIC 7 EXERCISES = 25 25 + 25 25 = 1+1 =2 Retain due to x2 of 2 < critical x2 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 rejecting H0. Accepting this would lead to a greater p value than significance level. Retaining H0 at the 5% level suggests that p value >a=0.05. *19.13 In 1912, over 800 passengers perished after the ocean liner Titanic collided with an iceberg and sank. The table below 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 Decision Rule: We will reject the null hypothesis at 0.05 level of significance if x2 > 3.84, given the degrees of freedom: = (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 TOPIC 7 EXERCISES fe (steerage, survived) = (712)(485) 1291 = 345320 1291 = 276.48 fe (cabin, not survived) = (579)(806) 1291 = 466674 1291 = 361.48 fe (steerage, not survived) = (712)(806) 1291 = 573872 1291 = 444.52 Survived/Accommodation s 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 TOPIC 7 EXERCISES A cabin passenger is 3.06 times more likely to survive than a steerage passenger. 19.14 In a classic study, Milgram et al. “lost” stamped envelopes with fictitious addresses (Medical Research Association, Personal Address, Friends of Communist Party, and Friends of Nazi Party).* One hundred letters with each address were distributed among four locations (shops, cars, streets, and phone booths) in New Haven Connecticut, with the following results: (a) Using the .05 level of significance, test the null [Show More]

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