Midterm_Problems_Solutions

# Midterm_Problems_Solutions - STA 100 Midterm I Practice...

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STA 100: Midterm I Practice Question Solutions For all questions you must show your working. This enables us to understand your thought process, give partial credit and prevent crude cheating. 1. Conditional Probabilities: You are part of a research lab investigating genetic factors inﬂuencing susceptibility to a class of rare bone diseases. Having done extensive research your lab has collected the following data on 256 subjects that it believes to be representative of the wider U.S. population: | Gene X Activated Gene X Inactivated | Total --------------------------------------------------------------------- Disease | 31 8 | 39 No Disease | 97 120 | 217 --------------------------------------------------------------------- Total | 128 128 | 256 Comments in brackets were not required, only the numeric answers are needed. (a) Using the above data, estimate the proportion of people who have the disease. 39/256 = 0.15 (Although this is the best we can do, this is probably a worthless estimate. It looks like the lab chose to sample equal numbers of people with Gene X and without Gene X, so the proportion who have the disease may not be properly weighted.) (b) Estimate the proportion of the people who have Gene X activated. 128/256 = 0.5. (Again, this is probably a worthless estimate though, since it looks like the lab chose to sample equal numbers of people with Gene X and without Gene X.) (c) Estimate the chance of a person having the disease if they have Gene X activated. 31/128 = 0.24 (This is more likely to be a plausible estimate, since it does not depend on the relative proportions of persons having Gene X or not.) (d) Estimate the chance of a person having Gene X activated if they have the disease. 31/39 = 0.79 (Again, this is not likely to be a reliable estimate, since it requires the correct weighting of Gene X/not Gene X.) (e) Which of the two previous probabilities is more relevant to your lab? Explain. The probability of someone having the disease given that they have Gene X activated is likely of more interest. The comparison of P ( Disease | Gene X ) and P ( Disease | No Gene X ) directly addresses the question of whether those with Gene X are more or less susceptible to the disease. The other direction, P ( Gene X | Disease ), depends on how many of the population have Gene X activated – this is arguably not of direct interest.

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6 8 10 12 14 Boxplot of Hamster/Gerbil Lengths Figure 1: Boxplot of hamster and gerbil lengths (in cm) 2. You are working for a local veterinarian, and encounter some data on the size of hamsters and gerbils sold by a local store. You have data on the length (in cm, head-to-tail) or each of 160 hamsters/gerbils. This data is displayed in a boxplot in Figure 1. (a) Describe what type of distribution this is. It is close to a symmetric distribution, not strongly right or left skewed. (b) The median for this data is 9.8 cm. Provide an estimate of the mean for this data. Roughly 9.8cm as well. For a symmetric distribution, the mean is roughly equal to the median.
6 10 12 14 0 2 4 6 8 10 Histogram of Hamster/Gerbil Lengths x Figure 2: Histogram of hamster and gerbil lengths (in cm) Figure 2b is a histogram of the same data.

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Midterm_Problems_Solutions - STA 100 Midterm I Practice...

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