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Chapter 7: Problem 31

Airport security The Transportation Security Administration \((\mathrm{TSA})\) is responsible for airport safety. On some flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers \(-12\) in first class and 64 in coach class. TSA officers selected an SRS of 10 passengers for screening. Let \(\hat{p}\) be the proportion of first-class passengers in the sample. (a) Is the 10\(\%\) condition met in this case? Justify your answer. (b) Is the Normal condition met in this case? Justify your answer.

Short Answer

Expert verified
(a) No, the 10% condition is not met. (b) No, the Normal condition is not met.

Step by step solution

01

Understanding the Context

The problem involves determining whether certain statistical conditions are met when selecting a Simple Random Sample (SRS) of 10 passengers from a flight with both first-class and coach-class passengers. There are 12 first-class and 64 coach-class passengers in total, making up a total of 76 passengers.
02

Recall of Conditions

There are two conditions to check: the 10% condition and the Normal condition. The 10% condition suggests that the sample size should be less than 10% of the population to ensure independence among selections. The Normal condition helps to ensure that the sampling distribution of the sample proportion can be approximated by a normal distribution, which requires that both the expected number of successes and failures be at least 10.
03

Checking the 10% Condition

The 10% condition states that the sample size should be less than 10% of the population size. Here, the sample size is 10, and the population size is 76. Calculate 10% of the population: \(0.10 \times 76 = 7.6\). Since 10 is more than 7.6, the 10% condition is not met.
04

Calculate Expected Values for the Normal Condition

We need to assess if the Normal condition is satisfied by checking the expected counts of first-class and coach-class passengers in the sample. The probability of selecting a first-class passenger, \(p\), is \(\frac{12}{76}\). Calculate \(np\) for first-class passengers where \(n\) is the sample size: \(np = 10 \times \frac{12}{76} \approx 1.58\). Also, calculate \(n(1-p)\) for coach passengers: \(n(1-p) = 10 \times \frac{64}{76} \approx 8.42\). Neither 1.58 nor 8.42 is at least 10, so the Normal condition is not met.

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Key Concepts

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

10% condition
The 10% condition is a crucial part of evaluating the validity of a statistical sample. It ensures the independence of the sample selections from a larger population.
When we take a simple random sample (SRS) from a population, the sample size should be less than 10% of the entire population. This guideline helps maintain that the selections are independent of each other.
If you imagine a deck of cards, the removal of one card can greatly impact the remaining deck if the removed card was too significant. Similarly, if a sample is too large relative to the population, it could affect the outcomes of the analysis.
  • The population is 76 passengers.
  • The sample should not exceed 7.6 passengers (10% of 76) to meet the condition.
  • In this case, the sample size is 10, which is more than 10% of the population.
This means the 10% condition is not fulfilled here.
Normal condition
The Normal condition allows statisticians to use the normal distribution as an approximation to model sample proportions. This approximation is important because it makes the data easier to analyze using statistical methods.
To satisfy the Normal condition, we check the expected counts of both successes and failures in our sample. Both should be at least 10.
  • Consider the proportion of first-class passengers: 12 out of 76.
  • Use this to predict how many first-class passengers could appear in a random sample of 10.
  • Calculate the expected count: \( np = 10 \times \frac{12}{76} \approx 1.58 \)
  • And the expected count of coach-class passengers: \( n(1-p) = 10 \times \frac{64}{76} \approx 8.42 \)
Neither number reaches 10, thus the Normal condition is not satisfied. This tells us that a normal distribution is not a good fit for this sample.
sampling distribution
Sampling distribution describes the probabilities of different outcomes for a sample statistic over all possible samples from a population.
It is a key concept when analyzing the variability and shape of the distribution of a sample proportion.
While dealing with sample proportions, understanding the sampling distribution is essential to determine the alignment with a normal distribution, thus impacting statistical inference. However, when conditions such as the Normal condition are not satisfied, the results could be misleading.
  • The shape of the sampling distribution depends on two factors: sample size and population proportion.
  • Even when attempting to apply a normal approximation through conditions, the distribution may not always align properly if conditions are unmet.
  • With our current example, the sample size is too small to satisfy the conditions needed for a reliable normal approximation.
Hence, sampling distribution analysis can become challenging without these conditions.
sample proportion
Sample proportion, denoted usually as \( \hat{p} \), represents the fraction of the sample that exhibits a particular trait.
It is a statistic used to estimate the corresponding population proportion \( p \). This measure is incredibly useful in statistics because it provides a straightforward way of generalizing findings from a small group to an entire population.
In this example, the sample proportion refers to the fraction of first-class passengers in a sample of 10 selected randomly.
  • Sample proportion \( \hat{p} \) is calculated by dividing the number of successes (first-class passengers) by the total sample size (10 passengers).
  • It's used to infer about the proportion of first-class passengers in the larger group of 76 passengers.
  • However, with insufficient sample size, as indicated by unmet conditions, this estimate may not be reliable.
Therefore, understanding the sample proportion is essential, but its accuracy is reliant on meeting certain statistical conditions.

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Most popular questions from this chapter

A newspaper poll reported that 73\(\%\) of respondents liked business tycoon Donald Trump. The number 73\(\%\) is (a) a population. (b) a parameter. (c) a sample. (d) a statistic. (e) an unbiased estimator.

Do you drink the cereal milk? AUSA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, 67\(\%\) said that they drink it. Suppose that 70\(\%\) of U.S. adults actually drink the cereal milk. Let \(\hat{p}\) be the proportion of people in the sample who drink the cereal milk. (a) What is the mean of the sampling distribution of \(\hat{p} ?\) Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p} .\) Check to see if the 10\(\%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Normal condition is met. (d) Find the probability of obtaining a sample of 1012 adults in which 67\(\%\) or fewer say they drink the cereal milk. Do you have any doubts about the result of this poll?

Run a mile During World War II, \(12,000\) able bodied male undergraduates at the University of Illinois participated in required physical training. Each student ran a timed mile. Their times followed the Normal distribution with mean 7.11 minutes and standard deviation 0.74 minute. An SRS of 100 of these students has mean time \(\overline{x}=7.15\) minutes. A second SRS of size 100 has mean \(\overline{x}=6.97\) minutes. After many SRSs, the values of the sample mean \(\overline{x}\) follow the Normal distribution with mean \(7.11 \mathrm{minutes}-\) and standard deviation 0.074 minute. (a) What is the population? Describe the population distribution. (b) Describe the sampling distribution of \(\overline{x} .\) How is it different from the population distribution?

Dem bones \((2.2)\) Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD score that is 2.5 standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and gender roughly follow a Normal distribution. (a) What percent of healthy young adults have osteoporosis by the WHO criterion? (b) Women aged 70 to 79 are, of course, not young adults. The mean BMD in this age group is about \(-2\) on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis?

A sample of teens A study of the health of teenagers plans to measure the blood cholesterol levels of an SRS of \(13-\) to 16 -year olds. The researchers will report the mean \(\overline{x}\) from their sample as an estimate of the mean cholesterol level \(\mu\) in this population. (a) Explain to someone who knows no statistics what it means to say that \(\overline{x}\) is an unbiased estimator of \(\mu\) (b) The sample result \(\overline{x}\) is an unbiased estimator of the population mean \(\mu\) no matter what size SRS the study chooses. Explain to someone who knows no statistics why a large random sample gives more trustworthy results than a small random sample.
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