Fire hydrant inspections

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

1. A.

   Suppose the water department's claim is true. Let $X$ be the number of fully functional hydrants in a random sample of 60 of the city's hydrants.

   i. The 60 hydrants are selected without replacement. Explain why a binomial distribution is still a reasonable model for $X$.

   ii. Calculate the mean and the standard deviation of $X$.

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   Score part A

   • 1 point Explains that the sample of 60 is no more than 10% of the 2,400 hydrants, so the selections are approximately independent even though they are made without replacement. The response may also note the other binomial conditions: a fixed number of trials, two outcomes, and a probability of 0.95 of success on each trial.Earns it: "60 is less than 10% of 2,400 (240), so removing each selected hydrant hardly changes the proportion left, and the trials are close to independent."Does not: "Each hydrant either works or does not." (true, but it does not address sampling without replacement); "Because the sample is random." (random selection alone does not make the trials independent); "Because $np=57\ge 10$." (that is part of a normality check, not a reason the binomial model applies)
   • 1 point Calculates ${\mu }_X=60(0.95)=57$ and ${\sigma }_X=\sqrt{60(0.95)(0.05)}\approx 1.69$ hydrants, with work shown.Earns it: "Mean $=60(0.95)=57$. SD $=\sqrt{60(0.95)(0.05)}=1.69$."Does not: "Mean 57, SD 2.85" (2.85 is the variance; the standard deviation is its square root); "Mean 0.95, SD 0.028" (these describe the sample proportion, not the count $X$)

   Model response

   i. The sample of 60 hydrants is less than 10% of the 2,400 hydrants ($60\le 240$), so removing each selected hydrant barely changes the proportion of functional hydrants that remain, and the 60 selections are close to independent. Each hydrant is either fully functional or not, $n=60$ is fixed, and the probability of selecting a fully functional hydrant is 0.95 each time, so $X$ is approximately binomial.

   ii. ${\mu }_X=np=60(0.95)=57$ hydrants, and ${\sigma }_X=\sqrt{np(1-p)}=\sqrt{60(0.95)(0.05)}\approx 1.69$ hydrants.

2. B.

   The council member wants to test whether the proportion of all the city's hydrants that are fully functional is less than 0.95.

   i. State the null and alternative hypotheses for the test, and define the parameter.

   ii. Explain why the normality condition for a one-sample z-test for a population proportion is not met.

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   Score part B

   • 1 point States $H_0:p=0.95$ and $H_a:p<0.95$, and defines $p$ as the proportion of all the city's hydrants that are fully functional.Earns it: "$H_0:p=0.95$, $H_a:p<0.95$, where $p$ is the proportion of all 2,400 city hydrants that are fully functional"Does not: "$H_a:p\ne 0.95$" (the council member suspects that fewer hydrants work, which is a one-sided alternative); "$H_0:\hat{p}=0.95$" (hypotheses are about the population proportion, not the sample proportion of 0.90); "$p$ is the proportion of the 60 hydrants that are functional" (that describes the sample)
   • 1 point Shows that the expected number of hydrants that are not fully functional, $60(1-0.95)=3$, is less than 10, so the normality condition is not met. Using the observed count of 6 hydrants that are not fully functional, which is also less than 10, also earns the point.Earns it: "$n(1-p_0)=60(0.05)=3$, which is less than 10."Does not: "The sample is too small." (does not identify which count falls below 10); "$60(0.95)=57\ge 10$" (checks only the expected successes, which pass); "The sample of 60 is more than 10% of the hydrants." (60 is only 2.5% of 2,400, and the 10% condition is a different condition)

   Model response

   i. $H_0:p=0.95$ and $H_a:p<0.95$, where $p$ is the proportion of all 2,400 of the city's fire hydrants that are fully functional.

   ii. The expected number of hydrants that are not fully functional is $n(1-p_0)=60(0.05)=3$, which is less than 10. The normality condition fails, so the sampling distribution of $\hat{p}$ may not be approximately normal and the z-test's p-value could not be trusted.

3. C.

   i. Use the simulation results in the table to estimate the p-value of the test. Show your work.

   ii. Interpret the p-value from (i) in context.

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   Score part C

   • 1 point Counts the simulated samples with 54 or fewer fully functional hydrants, $4+6+18+50=78$ of 1,000, and estimates the p-value as $\frac{78}{1000}=0.078$.Earns it: "$(4+6+18+50)/1000=0.078$"Does not: "28/1000 = 0.028" (counts only samples with fewer than 54, leaving out the samples equal to the observed result); "50/1000 = 0.05" (counts only samples with exactly 54); "922/1000 = 0.922" (counts the wrong tail)
   • 1 point Interprets the p-value as the probability, computed assuming the null hypothesis is true (95% of all the city's hydrants are fully functional), of getting 54 or fewer fully functional hydrants in a random sample of 60. The assumption that the null hypothesis is true and the "or fewer" tail must both be present.Earns it: "If 95% of all the city's hydrants are fully functional, there is about a 0.078 probability that a random sample of 60 would have 54 or fewer fully functional hydrants."Does not: "There is a 7.8% chance that 95% of the hydrants are functional." (the p-value is not the probability that the null hypothesis is true); "The probability of getting exactly 54 functional hydrants is 0.078." (leaves out the "or fewer" tail and the assumption); "7.8% of the city's hydrants do not work." (the p-value is a probability about samples, not a proportion of hydrants)

   Model response

   i. In the simulation, $4+6+18+50=78$ of the 1,000 samples had 54 or fewer fully functional hydrants, so the estimated p-value is $\frac{78}{1000}=0.078$.

   ii. Assuming that 95% of all the city's hydrants are fully functional, the probability of getting 54 or fewer fully functional hydrants in a random sample of 60 hydrants is about 0.078.

4. D.

   Using a significance level of $\alpha =0.05$, justify a conclusion in context.

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   Score part D

   • 1 point Compares the p-value with the significance level and makes a formal decision about the null hypothesis: because $0.078>0.05$, fail to reject $H_0$. The p-value used in the comparison must be correct.Earns it: "0.078 > 0.05, so we fail to reject $H_0$."Does not: "0.078 > 0.05, so reject $H_0$." (the decision does not match the comparison); "0.078 > 0.05, so we accept $H_0$." (a test can fail to reject the null hypothesis but can never show it is true); "0.028 < 0.05, so reject $H_0$." (uses an incorrect p-value)
   • 1 point States that there is not convincing statistical evidence that the proportion of all the city's hydrants that are fully functional is less than 0.95. The conclusion must be in terms of the alternative hypothesis and refer to the population proportion in context.Earns it: "There is not convincing statistical evidence that the proportion of all the city's hydrants that are fully functional is less than 0.95."Does not: "The data prove that 95% of the hydrants are fully functional." (failing to reject is not evidence that the null hypothesis is true); "There is convincing evidence that fewer than 95% of the hydrants work." (does not match the decision)

   Model response

   Because the p-value of 0.078 is greater than $\alpha =0.05$, we fail to reject $H_0$. There is not convincing statistical evidence that the proportion of all the city's fire hydrants that are fully functional is less than 0.95.

5. E.

   i. Based on your conclusion in part D, identify the type of error that could have been made, and describe that error in context.

   ii. Before the inspections, the council member considered using a significance level of 0.10 instead of 0.05. Explain how using 0.10 would change the probabilities of Type I and Type II errors, and explain why the council member might prefer 0.10 in this situation.

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   Score part E

   • 1 point Identifies a Type II error and describes it in context: the proportion of all the city's hydrants that are fully functional really is less than 0.95, but the test does not find convincing evidence of it. An error type consistent with an incorrect decision in part D also earns the point if it is described correctly.Earns it: "Type II error: in reality fewer than 95% of the city's hydrants work, but the test did not find convincing evidence of that."Does not: "Type I error" (a Type I error can happen only when the null hypothesis is rejected); "Type II error" with no description in context; "Type II: concluding that 95% work when 95% really do work" (that describes a correct decision)
   • 1 point Explains that a significance level of 0.10 increases the probability of a Type I error and decreases the probability of a Type II error, and gives a reason in context for preferring it, such as: failing to notice that too many hydrants do not work could be dangerous in a fire, while a false alarm would only lead to extra inspections or repairs.Earns it: "Raising alpha to 0.10 makes a Type I error more likely and a Type II error less likely. Missing broken hydrants is the dangerous mistake, because firefighters could arrive at a hydrant that does not work."Does not: "0.10 makes a Type II error more likely." (a larger significance level makes rejecting easier, so a Type II error becomes less likely); "0.10 is better because the result would then be significant." (a significance level must be chosen before seeing the data, not to reach a desired result)

   Model response

   i. A Type II error. It is possible that the proportion of all the city's hydrants that are fully functional really is less than 0.95, but the test did not find convincing evidence of it, so the problem would go unnoticed.

   ii. A significance level of 0.10 makes it easier to reject $H_0$, so the probability of a Type I error increases and the probability of a Type II error decreases. The council member might prefer 0.10 because a Type II error here means not noticing that too many hydrants do not work, which could put lives and property at risk in a fire. A Type I error would only lead to extra inspections or repairs that turned out not to be needed.

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