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Chapter 8: Problem 118

Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. a. i. \(x=\) _____ ii. \(n=\) _____ iii. \(p^{\prime}=\) _____ b. Define the random variables \(X\) and \(P^{\prime},\) in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 95\(\%\) confidence interval for the population proportion who claim they always buckle up. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. e. If this survey were done by telephone, list three difficulties the companies might have in obtaining results.

Short Answer

Expert verified
i. \( x = 320 \), ii. \( n = 400 \), iii. \( p^{\prime} = 0.8 \). Random variables: \( X \) number claiming to always buckle up, \( P^{\prime} \) sample proportion claiming to. Use Normal approximation. 95% CI: (0.7608, 0.8392). Challenges: non-response bias, selection bias, response bias.

Step by step solution

01

Identify the given values

i. In part a(i), \( x \) represents the number of drivers who claimed they always buckle up, which is given as 320. ii. In part a(ii), \( n \) is the total number of drivers surveyed, which is 400. iii. \( p^{\prime} \) is the sample proportion and can be calculated as \( \frac{x}{n} = \frac{320}{400} = 0.8 \).
02

Define the random variables

b. The random variable \( X \) represents the number of drivers in the survey who claim they always buckle up. The random variable \( P^{\prime} \) (read as "p-prime") is the sample proportion, i.e., the proportion of drivers in the sample who claim they always buckle up.
03

Choose an appropriate distribution

c. Since we are interested in constructing a confidence interval for a population proportion and the sample size is large, we should use the Normal approximation to the Binomial distribution. This is appropriate when both \( np \) and \( n(1-p) \) are greater than 5, which is satisfied here.
04

Construct a 95% confidence interval

d. To construct a 95% confidence interval for the population proportion: i. The confidence interval is calculated by:\[ p^{\prime} \pm Z \cdot \sqrt{\frac{p^{\prime}(1-p^{\prime})}{n}} \]Given \( Z = 1.96 \) for a 95% confidence level, and \( p^{\prime} = 0.8 \), \( n = 400 \), \[ E = 1.96 \cdot \sqrt{\frac{0.8 \cdot 0.2}{400}} = 0.0392 \]Thus, the confidence interval is \( 0.8 \pm 0.0392 \), or \( (0.7608, 0.8392) \). ii. Sketching the graph involves drawing a normal curve centered at 0.8 with the endpoints of the interval at 0.7608 and 0.8392. iii. The error bound, \( E \), is 0.0392.
05

Discuss potential survey difficulties

e. Difficulties in obtaining results via telephone may include: 1) Non-response bias, where people may not answer or refuse to participate; 2) Selection bias, if certain demographics are more likely to be reached by phone; 3) Response bias, where respondents may not provide truthful answers over the phone.

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Key Concepts

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

Understanding Population Proportion
A population proportion represents a fraction of the entire population that exhibits a certain characteristic. In this problem, we're interested in the population proportion of drivers who claim to always buckle up. A survey was conducted, and through this survey, we derived a sample proportion, also known as the sample statistic. This is a smaller group taken from the larger population to estimate the population proportion. For example, if 320 out of 400 drivers in our sample claim they always buckle up, our sample proportion is 0.8. This figure helps infer how widespread a trait might be in the total population, giving us insight without needing to survey everyone.
Normal Approximation: Simplifying Calculations
When dealing with large sample sizes, the binomial distribution鈥攁 type of distribution applied when there are two possible outcomes鈥攃an be approximated to a normal distribution. This is especially useful because the normal distribution is mathematically easier to work with.For a normal approximation to be valid, certain conditions must be met. Specifically, both \(np\) and \(n(1-p)\) should be greater than 5, where \(n\) is the sample size and \(p\) is the sample proportion. In our case, these conditions are fulfilled, allowing us to use the normal approximation. This means confidence intervals, which help us understand the range within which the true population proportion lies with a certain degree of confidence, can be calculated with relative ease using normal distribution principles.
The Role of Binomial Distribution
The binomial distribution is applicable when we're interested in the number of successes in a fixed number of trials, each with the same probability of success. In other words, it helps model situations where there are clear yes/no outcomes, like our survey on buckling up. Here, each driver interviewed either buckles up all the time鈥攁 success鈥攐r doesn't. Being naturally aligned with how the survey collects data, the binomial distribution is an excellent starting point for analysis. However, due to complex calculations, in large samples, we often switch to using the normal approximation to make life easier while still achieving accurate results. This is why, although the binomial distribution is theoretically used, we substituted it with normal approximation for constructing our confidence intervals.
Why Consider Survey Bias?
Survey bias refers to various factors that can skew the results of a survey, making them unrepresentative of the true population. When conducting a survey via telephone, there are several types of bias to consider:
  • Non-response bias: Some individuals may choose not to answer their phone or participate, leading to a sample that's not reflective of the entire population.
  • Selection bias: If only certain groups of people are reachable by phone鈥攕ay, older individuals or those with landlines鈥攜our sample may be skewed towards those groups.
  • Response bias: Respondents may not always give truthful responses, especially in surveys where social desirability can influence their answers.
Being aware of these biases helps in interpreting survey results with a grain of caution, ensuring we understand the potential limitations and inaccuracies in our data collection methods.

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Most popular questions from this chapter

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Define the random variable \(X\) in words.

Use the following information to answer the next 13 exercises: The data in Table 8.10 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. $$\begin{array}{|c|c|}\hline X & {\text { Freq }} \\ \hline 1 & {1} \\\ \hline 2 & {7} \\ \hline 3 & {78} \\ \hline 4 & {7} \\ \hline 5 & {6} \\\ \hline\end{array}$$ What is \(\overline{x}\) estimating?

Use the following information to answer the next five exercises: of \(1,050\) randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks. Suppose we want to lower the sampling error. What is one way to accomplish that?

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. Which distribution should you use for this problem?
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