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5 Cards in this Set
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A student interested in comparing the effect of different types of music on short-term memory conducted the following study: 80 volunteers were randomly assigned to one of two groups. The first group was given 5 minutes to memorize a list of words while listening to rap music. The second group was given the same task while listening to classical music. The number of words correctly recalled by each individual was then measured, and the results for the two groups were compared. A. Is this an experiment or an observational study? Justify your answer.
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This is an experiment, because a treatement (type of music) was imposed on the subjects.
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A student interested in comparing the effect of different types of music on short-term memory conducted the following study: 80 volunteers were randomly assigned to one of two groups. The first group was given 5 minutes to memorize a list of words while listening to rap music. The second group was given the same task while listening to classical music. The number of words correctly recalled by each individual was then measured, and the results for the two groups were compared. B. In the context of this study, explain why it is important that the subjects were randomly assigned to the two experimental groups (rap music and classical music).
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We randomly assign subjects to groups in order to create two groups that are as similar as possible with respect to any variables that might influence the subjects' capacity for recalling words. That way, any differences we see in the mean number of words recalled can be attributed to either the type of music or to variation arising from random assignment. For example, if subjects were not assigned at random and were allowed to choose which group, people who are easily distracted and may have more difficulty memorizing a list of words may tend to choose classical music because there are usually no lyrics that would be distracting.
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We would like to estimate the mean employee's salary at a firm with 99% confidence interval to an accuracy level within $2,000. When the actual survey was conducted the salary of $6,000 was the actual employee average. How many people will need to be surveyed to meet the confidence interval?
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The 99% confidence level translates to a z or t of 2.58. Therefore, the sample size would be:
2,000= (2.58*6,000)/√n n=((2.58*6,000)/2000)² n=59.9 |
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The school was trying to determine the number of students that wear eye glasses. The margin of error was 3.5 with a confidence interval of 95%. The prior judgement was 30%. What will be the minimum sample size to reach the margin of error level?
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ME represents margin of error.
ME=z*√[p(1-p)/n] where z is z-score, p is prior judgement and n is the sample size. ME=z*√[p(1-p)/n] 1.96=√[0.3(1-0.3)/n] n=660 The sample size needed is 660 people |
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Alan is trying to determine that fast food consumption results in people being heavy. People from house 1 were eating fast food regularly. People from house 2 did not eat fast food. Based on the data, is there any reason to believe that people from house 1 weighed more? House 1 weights(kg): 30, 48, 55, 34, 41, 47, 59, 50, 39, 35, 85, 95, 75 House 2 weights (kg): 74, 53, 94, 82, 77, 82, 84, 47, 55, 80, 75, 90, 78, 62, 71, 79
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There is no reason to believe that people from house 1 weighed more from the data. When we look at these values: Home 1: SD=21.1, Mean=57.8, Median=52.5 Home 2: SD=13.3, Mean=73.93, Median=77.5, Home 2 is almost 16 units greater on average, and the group is more closely distributed with a smaller standard deviation.
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