soci-301-assignment 6

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Athabasca University, Athabasca *

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301

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Sociology

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Jan 9, 2024

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lOMoARcPSD|35329777 Question 1: In a particular country, it is known that college students report falling in love an average of 2.20 times during their college years. A sample of five college students originally from that country but who have spent their entire college career in Canada, were asked how many times they had fallen in love during their college years. Their numbers were 2, 3, 5, 5, and 2. Using the .05 significance level, do exchange students like these who go to college in Canada fall in love more often than those who complete their studies in their country of origin? Use the hypothesis–testing steps to answer this. Step 1: To examine if students who attend college in Canada fall in love more frequently than those who finish their education in their home country. Null hypothesis: Ho: Students who attend college in Canada do not fall in love as much as those who finish their studies in their own country.. mHo 2.2 The research hypothesis: H1:Students who attend college in Canada are more likely to fall in love than those who finish their education in their home country. mH1>2.2 The level of significance is 0.5 (95%) N=5 m= 2.2 One-tailed Test Step 2: Figure estimated population variance of different scores mean of x = ∑(x)/n = (2+3+5+5+2)/5 = 17/5 = 3.4 Df = n-1 = 5-1 = 4 Estimated population standard deviation: S 2= ∑ {X-M} 2/5-1 = ∑{(2-2.2)+(3-2.2)+(5-2.2)+(5-2.2)+(2-2.2)}2 / 4 = 16.4 /4 = 4.1 S =√4.1 = 2.02 Step 3: Figure SD of distribution of means of different scores

lOMoARcPSD|35329777 The variance of the mean distribution is calculated by multiplying the predicted population variance by the number of scores in the sample: S M=√ (4.1/ 5) = √ 0.82= 0.9 Step 4: Determine cutoff sample t score: cutoff t score in t-test table = 2.13 (Need to reject the null hypothesis if sample t score > 2.13 in t test because it’s too extreme to accept the null hypothesis) Step 5: Conclusion for null hypothesis: T-test : t = (3.4 – 2.2) / 0.9 = 1.33 The sample t score of 1.33 is lower than the threshold t score of 2.13, based on the results. To validate the research hypothesis, accept the null hypothesis. This decision doesn’t support the research hypothesis. Question 2: Five people who were convicted of speeding were ordered by the court to attend a workshop. A special device placed in their cars kept a record of their speeds for 2 weeks before and after the workshop. The maximum speeds for each person during the 2 weeks before and the 2 weeks after the workshop follow: Using the .05 significance level, should we conclude that people are likely to drive more slowly after this workshop? Use hypothesis testing. Two tailed t test, two data sets for each identical participants. Participant Before After Difference Deviation (Difference - M ) Squared Deviation

lOMoARcPSD|35329777 H0: d 0 H1: d < 0 N= 5 M= -20/ 5= - 4 Step 1: Figure estimated sample standard deviation of different score: S2= =114/ (5-1) = 28.5 SD= √28.5 = 5.33 Step 2: Figure variance of distribution of means of different scores S2 M =S2/N =28.5/5 =5.7 Step 3: Figure SD of distribution of means of different scores S M =√5.7 = 2.39 Step 4: Cutoff t score? (two-tailed test) cutoff T score with df: 4 needed for 5% significance level: 2.78 Step 5: T-test and 58 49 L.B 65 -7 -7 49 J.K 62 65 3 7 60 56 -4 R.C 0 0 66 R.T 70 -4 0 0 60 J.M -8 68 -4 16 = -20 = 114 Hypothesis:

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lOMoARcPSD|35329777 determine sample t score on the comparison distribution sample t score = (M-population M) / 2.78 = -4-0 / 2.78 = -1.44 Decision: -1.44 < 2.78 This result doesn’t reject Null hypothesis At the 95 percent significance level, this decision does not support the study hypothesis. After this workshop, people are unlikely to drive more slowly. Question 3: Five university students were given an English achievement test before and after receiving instruction in basic grammar. Their scores are shown below. Hypothesis: H0: d 0 H1: d < 0 Mean of difference in each student: (18-20)+(22-18)+(15-17)+(17-16)+(9-12)/5 = (-2+4-2+1-3)/5 = - 0.4 Figure Sample standard deviation of difference: df=4 SD2= = 8.5

lOMoARcPSD|35329777 S2 M =S2/N = 8.5 /5= 1.7 S M= √ 1.7= 1.3 Significance level is 0.05 Critical t cutoff score in one-tailed table is 2.13 Need to reject null hypothesis If sample t score is > 2.13 t- test: t= M-population M / SM = -4-0/1.3 = -3.08 Decision: sample t score is smaller than cutoff t score, The null hypothesis is not rejected, and the research hypothesis is not supported. Following grammar instruction, future pupils are unlikely to increase their English results. Question 4: Figure the estimated effect size and indicate whether it is approximately small, medium or large for each of the following studies. Estimate effect size: M/S (a) 8/30=0.267 According to the effect table, 0.267 is closer to 0.20 so this effect size is small (b) 8/10=0.80 According to the effect table, 0.80 is the size of large effect size, .80. (c) 16/30= 0.53 According to the effect table, 0.53 is closer to medium size of 0.5

lOMoARcPSD|35329777 Question 5: What is the power of the following studies, using a t-test for dependent means (based on the . 05 significance level)? (a) Using the provided table in the textbook page 260 in Chapter 8, an effect size of small with differencescores in the sample is 50, and two tails the power is 0.29 (b) Using the provided table in the textbook page 260 in Chapter 8, an effect size of medium with difference scores in the sample is 50, and two tails the power is 0.94

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