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A city ordinance requires that a smoke detector be installed in all residential housing. There is concern that too many residences are still without detectors, so a costly inspection program is being contemplated. Let \(\pi\) be the proportion of all residences that have a detector. A random sample of 25 residences is selected. If the sample strongly suggests that \(\pi<.80\) (less than \(80 \%\) have detectors), as opposed to \(\pi \geq .80\), the program will be implemented. Let \(x\) be the number of residences among the 25 that have a detector, and consider the following decision rule: Reject the claim that \(\pi=.8\) and implement the program if \(x \leq 15\). a. What is the probability that the program is implemented when \(\pi=.80\) ? b. What is the probability that the program is not implemented if \(\pi=.70 ?\) if \(\pi=.60 ?\) c. How do the "error probabilities" of Parts (a) and (b) change if the value 15 in the decision rule is changed to 14 ?

Short Answer

Expert verified
The answers would be the calculated probabilities for each part. Note that actual values would depend on the specific calculations using the binomial probability formula and could be computed using either binomial probability tables, statistical software, or a scientific calculator that has the capability for these specific computations.

Step by step solution

01

Compute probability for part (a)

Compute the binomial probability \(P(x \leq 15)\) when \(\pi = 0.80\). This can be done using the binomial probability formula \[ P(x) = C(n,x) \times (π^x) \times ((1-π)^(n-x)) \]where \(n=25\),\(x=15\), and \(\pi = 0.80\). The cumulative of all the probabilities when \(x\) is 0 to 15 should be calculated as we are interested in the event of \(x \leq 15\)
02

Compute probability for part (b)

Similarly, compute the binomial probabilities \(P(x > 15)\) when \(\pi = 0.70\) and when \(\pi = 0.60\). These probabilities represent the probabilities that the program is not implemented. These can be calculated by subtracting the cumulative probabilities of \(x\) from 0 to 15 from 1 as we are interested in the probabilities when \(x\) is larger than 15.
03

Adjust and recompute for part (c)

With the new decision rule of rejecting the null hypothesis if \(x \leq 14\), we should compute the probabilities in the same manners as part (a) and (b), but in this step, we should calculate cumulative probabilities of \(x\) from 0 to 14 for both \(\pi = 0.80\), \(\pi = 0.70\) and \(\pi = 0.60\). This will show how the 'error probabilities' change when we adjust the decision rule's threshold.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Statistical Hypothesis Testing
Statistical hypothesis testing is a fundamental procedure in statistics, used to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. In the context of the provided exercise, the null hypothesis (\( H_0 \)) is that at least 80% of residences have smoke detectors (\( \( \( \pi = 0.80 \) \) \)), while the alternative hypothesis (\( H_1 \) is that less than 80% have smoke detectors (\( \pi < 0.80 \)).

The decision to implement an inspection program based on a sample of 25 residences is an example of hypothesis testing. If the sample data (\( x \)), the number of residences with detectors, is less than or equal to 15, the null hypothesis is rejected, indicating that the program should be implemented. This process involves calculating binomial probabilities to make a data-driven decision.
Error Probabilities
Error probabilities in statistical hypothesis testing refer to the likelihood of making incorrect decisions based on sample data. There are two types of errors to consider:
  • Type I Error: The probability of rejecting the null hypothesis when it is actually true (\( \alpha \)). In the exercise, this is the probability of implementing the program when in fact at least 80% of residences already have detectors.
  • Type II Error: The probability of failing to reject the null hypothesis when the alternative hypothesis is true (\( \beta \)). This would be not implementing the program when in reality, less than 80% of residences have detectors.
Calculating these probabilities helps to assess the risks involved in the decision-making process, by trying to minimize these errors, a more reliable decision can be made regarding the inspection program.
Cumulative Probability
Cumulative probability is the probability that a random variable is less than or equal to a certain value. It accumulates the probabilities for all values below or at a given point. In the context of the exercise, the cumulative probability is used to find the likelihood that the number of residences with detectors (\( x \) falls within a certain range, such as 15 or less.

Cumulative probabilities in binomial distributions, like this scenario, are calculated by summing individual binomial probabilities from the lowest value of x up to the threshold value of x. This provides a comprehensive measure of how likely it is that the observed outcome falls within a certain range, which is vital for making the decision on whether to implement the inspection program.
Random Sampling
Random sampling is a statistical method used to select a subset of individuals from a larger population in such a way that each individual has an equal chance of being chosen. It is a key component of statistical hypothesis testing because it ensures that the sample is representative of the population. In our exercise, a random sample of 25 residences is taken to estimate the proportion (\( \pi \) of homes with smoke detectors.

The accuracy of the hypothesis test's conclusion is highly dependent on the randomness of the sample. If the sample is not truly random, it may lead to biased results and a higher likelihood of committing Type I or Type II errors. Thus, ensuring a truly random selection process is crucial for the reliability of the testing results and the subsequent decisions made about the inspection program.

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Most popular questions from this chapter

The Wall Street Journal (February 15,1972 ) reported that General Electric was sued in Texas for sex discrimination over a minimum height requirement of \(5 \mathrm{ft}\) 7 in. The suit claimed that this restriction eliminated more than \(94 \%\) of adult females from consideration. Let \(x\) represent the height of a randomly selected adult woman. Suppose that \(x\) is approximately normally distributed with mean 66 in. (5 ft 6 in.) and standard deviation 2 in. a. Is the claim that \(94 \%\) of all women are shorter than \(5 \mathrm{ft}\) 7 in. correct? b. What proportion of adult women would be excluded from employment as a result of the height restriction?

Because \(P(z<.44)=.67,67 \%\) of all \(z\) values are less than \(.44\), and \(.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91 st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the 30 th \(z\) percentile?

Four people \(-\mathrm{a}, \mathrm{b}, \mathrm{c}\), and \(\mathrm{d}-\) are waiting to give blood. Of these four, a and b have type AB blood, whereas c and d do not. An emergency call has just come in for some type AB blood. If blood samples are taken one by one from the four people in random order for blood typing and \(x\) is the number of samples taken to obtain an \(\mathrm{AB}\) individual (so possible \(x\) values are 1,2 , and 3 ), what is the probability distribution of \(x\) ?

Determine the value of \(z^{*}\) such that a. \(-z^{*}\) and \(z^{*}\) separate the middle \(95 \%\) of all \(z\) values from the most extreme \(5 \%\) b. \(-z^{*}\) and \(z^{*}\) separate the middle \(90 \%\) of all \(z\) values from the most extreme \(10 \%\) c. \(-z^{*}\) and \(z^{*}\) separate the middle \(98 \%\) of all \(z\) values from the most extreme \(2 \%\) d. \(-z^{*}\) and \(z^{*}\) separate the middle \(92 \%\) of all \(z\) values from the most extreme \(8 \%\)

Consider babies born in the "normal" range of \(37-\) 43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth wcight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\) ("Are Babies Normal?" The American Statistician [1999]: \(298-302\) ). a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

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