Phone cases and cracked screens

Use the given information to respond to parts A, B, C, D, E, and F. Label any subparts (e.g., i and ii) that may be present.

1. A.

   For each type of case, calculate the proportion of owners in the sample whose screen has cracked. Show your work.

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   Score part A

   • 1 point Calculates all three conditional proportions: $\frac{47}{118}\approx 0.398$ for no case, $\frac{85}{283}\approx 0.300$ for a slim case, and $\frac{27}{199}\approx 0.136$ for a rugged case.Earns it: "No case: 47/118 = 0.398. Slim: 85/283 = 0.300. Rugged: 27/199 = 0.136."Does not: "47/159, 85/159, 27/159" (these show which cases the owners of cracked screens used; the question asks for the proportion of each case group with a crack); "47/600, 85/600, 27/600" (these are joint proportions of all 600 owners, not proportions within each case type)

   Model response

   No case: $\frac{47}{118}\approx 0.398$. Slim case: $\frac{85}{283}\approx 0.300$. Rugged case: $\frac{27}{199}\approx 0.136$.

2. B.

   Identify a graphical representation that would be appropriate for comparing the distribution of screen condition across the three types of case.

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   Score part B

   • 1 point Identifies a display for two categorical variables: a segmented (stacked) bar chart, a side-by-side bar chart, a mosaic plot, or a separate pie chart for each type of case.Earns it: "a segmented bar chart"; "a side-by-side bar chart of relative frequencies"; "a mosaic plot"Does not: "a histogram" (histograms display a quantitative variable); "a boxplot" or "a scatterplot" (displays for quantitative variables); "a pie chart of the types of case" (one pie chart shows only one variable)

   Model response

   A segmented bar chart, with one bar for each type of case divided into the proportions of owners whose screens have and have not cracked.

3. C.

   Based on your answers to part A, determine whether there is an association between type of case and screen condition for the owners in the sample. Justify your answer.

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   Score part C

   • 1 point Determines that there is an association between type of case and screen condition for the owners in the sample.Earns it: "Yes, there is an association."Does not: "There is no association." (the proportions with cracked screens in part A are clearly different); "It cannot be determined without a significance test." (the question asks about the owners in the sample, which the proportions describe directly)
   • 1 point Justifies the determination by comparing the conditional proportions from part A: the proportion of owners with a cracked screen differs across the types of case (about 0.398, 0.300, and 0.136). Conditional proportions consistent with those reported in part A also earn the point.Earns it: "The proportion with cracked screens falls from 0.398 with no case to 0.300 with a slim case to 0.136 with a rugged case. These differ, so the variables are associated."Does not: "More slim-case owners had cracked screens (85) than owners with no case (47)." (compares counts, but the groups have different sizes); "The numbers in the table are different." (does not compare proportions)

   Model response

   Yes, there is an association. The proportion of owners whose screen has cracked is quite different for the three types of case: about 0.398 with no case, 0.300 with a slim case, and 0.136 with a rugged case. If there were no association, these three proportions would be about the same.

4. D.

   One owner is selected at random from the 600 owners in the sample.

   i. Calculate the probability that the owner uses a slim case and has a cracked screen.

   ii. Calculate the probability that the owner uses a rugged case or has a cracked screen.

   iii. Given that the owner's screen has cracked, calculate the probability that the owner uses no case.

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   Score part D

   • 1 point Calculates $P(\text{slim and cracked})=\frac{85}{600}\approx 0.142$.Earns it: "85/600 = 0.142"Does not: "85/283 = 0.300" (a conditional probability, not the probability of both events); "$\frac{283}{600}\cdot \frac{159}{600}\approx 0.125$" (multiplying assumes the events are independent, and they are not)
   • 1 point Calculates $P(\text{rugged or cracked})=\frac{199+159-27}{600}=\frac{331}{600}\approx 0.552$.Earns it: "(199 + 159 - 27)/600 = 331/600 = 0.552"Does not: "(199 + 159)/600 = 0.597" (counts the 27 owners with a rugged case and a cracked screen twice); "27/600 = 0.045" (that is the probability of both events, not either one)
   • 1 point Calculates $P(\text{no case}|\text{cracked})=\frac{47}{159}\approx 0.296$.Earns it: "47/159 = 0.296"Does not: "47/118 = 0.398" (this is the probability of a cracked screen given no case, the reverse condition); "47/600 = 0.078" (not restricted to owners whose screens cracked)

   Model response

   i. $P(\text{slim and cracked})=\frac{85}{600}\approx 0.142$

   ii. $P(\text{rugged or cracked})=\frac{199+159-27}{600}=\frac{331}{600}\approx 0.552$

   iii. $P(\text{no case}|\text{cracked})=\frac{47}{159}\approx 0.296$

5. E.

   i. Determine whether the events "uses a rugged case" and "screen has cracked" are mutually exclusive. Justify your answer.

   ii. If type of case and screen condition were independent, how many of the 600 owners would be expected to use no case and have a cracked screen? Show your work.

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   Score part E

   • 1 point Determines that the events are not mutually exclusive because they can occur together: 27 owners used a rugged case and had a cracked screen, so $P(\text{rugged and cracked})=\frac{27}{600}=0.045\ne 0$.Earns it: "Not mutually exclusive, because 27 owners have both a rugged case and a cracked screen, so P(both) = 0.045, not 0."Does not: "Mutually exclusive, because rugged cases protect screens." (27 owners had a rugged case and a cracked screen, so the events can happen together); "Not mutually exclusive, because $P(\text{cracked}|\text{rugged})$ is not equal to $P(\text{cracked})$." (that checks independence, not mutual exclusivity)
   • 1 point Calculates the expected count under independence as $600\cdot \frac{118}{600}\cdot \frac{159}{600}=\frac{118\cdot 159}{600}\approx 31.3$ owners.Earns it: "$600(118/600)(159/600)=31.3$ owners"; "$(118)(159)/600\approx 31.27$"Does not: "47" (the observed count, not the count expected under independence); "$\frac{118}{600}\cdot \frac{159}{600}\approx 0.052$" (a probability, not a number of owners)

   Model response

   i. The events are not mutually exclusive. 27 owners used a rugged case and had a cracked screen, so $P(\text{rugged and cracked})=\frac{27}{600}=0.045$, which is not 0. Both events can happen to the same owner.

   ii. Under independence, $P(\text{no case and cracked})=\frac{118}{600}\cdot \frac{159}{600}$, so the expected number of owners is $600\cdot \frac{118}{600}\cdot \frac{159}{600}=\frac{118\cdot 159}{600}\approx 31.3$.

6. F.

   A company that makes rugged cases says these data prove that rugged cases prevent cracked screens. Explain why the data do not support a cause-and-effect conclusion.

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   Score part F

   • 1 point Explains that the data come from an observational study (a survey) in which owners chose their own cases rather than being randomly assigned one, so other differences between owners could explain the association. Naming a plausible confounding variable is not required, but a vague statement such as "correlation is not causation" with no reference to how the data were collected does not earn the point.Earns it: "This was a survey, so case types were not randomly assigned. People who buy rugged cases might be more careful with their phones, and that could explain why fewer of their screens cracked."Does not: "Correlation does not mean causation." (does not explain why for this study); "600 owners is not enough data." (sample size does not determine whether a causal conclusion is justified)

   Model response

   The magazine conducted a survey, which is an observational study. Owners chose their own cases instead of being randomly assigned a type of case. Owners who choose rugged cases may differ in other ways, such as being more careful with their phones, and that could explain why fewer of them have cracked screens. Without random assignment, the data show an association but cannot show that rugged cases prevent cracks.

Your score: 0 of 10. Each point is earned on its own, the way the exam scores it.