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Chapter 3: Problem 8

Elderly drivers. In January 2011 , The Marist Poll published a report stating that \(66 \%\) of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was \(3 \%\) using a \(95 \%\) confidence level. \({ }^{32}\) (a) Verify the margin of error reported by The Marist Poll. (b) Based on a \(95 \%\) confidence interval, does the poll provide convincing evidence that more than \(70 \%\) of the population think that licensed drivers should be required to retake their road test. once they turn \(65 ?\)

Short Answer

Expert verified
The margin of error is approximately 2.9%, verifying the report, but the poll does not support the claim that over 70% agree.

Step by step solution

01

Determine the Sample Proportion

The given sample proportion \( \hat{p} \) is 66%. Convert 66% to a decimal by dividing by 100: \( \hat{p} = 0.66 \).
02

Determine the Standard Error

The standard error (SE) for a proportion is calculated with the formula: \[ SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \] where \( n \) is the sample size of 1018. Plug in the values: \[ SE = \sqrt{ \frac{0.66 \times 0.34}{1018} } \approx 0.0148 \].
03

Calculate the Margin of Error

The margin of error (ME) is calculated with the formula: \[ ME = z \times SE \] where \( z \) for a 95% confidence level is approximately 1.96. Thus: \[ ME = 1.96 \times 0.0148 \approx 0.029 \] or 2.9%. The reported 3% margin is due to rounding.
04

Construct the Confidence Interval

The 95% confidence interval is calculated using: \[ CI = \hat{p} \pm ME \] Plug in our values: \[ CI = 0.66 \pm 0.029 \] which gives the interval \( (0.631, 0.689) \).
05

Analyze the Confidence Interval for Hypothesis Testing

We need to check if the interval provides evidence that more than 70% support retesting. Since the interval \((0.631, 0.689)\) lies entirely below 0.70, it does not support the claim that more than 70% of the population agrees.

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Key Concepts

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

Margin of Error
In statistics, the margin of error provides an estimate of the range within which the true population parameter lies, given a certain confidence level. It helps us understand the precision of our sample estimate. The margin of error is usually associated with the confidence interval, which indicates that the true parameter is within this range a certain percentage of the time.

To calculate the margin of error, we use the formula:
  • \( ME = z \times SE \)
Here, "\( z \)" is the z-score corresponding to the desired confidence level. For a 95% confidence level, which is most frequently used, the z-score is approximately 1.96.
The standard error (SE) represents the expected variation in the sample proportion. Multiplying it by the z-score gives us the margin of error. In the exercise example, the margin of error was calculated as 2.9%, close to the reported 3%, showcasing the slight impacts of rounding in real-world reporting.

In summary, the margin of error accounts for random sampling errors and gives us a range in which the true value likely falls, helping us understand the reliability of the sample estimate.
Standard Error
The standard error (SE) measures how much variability or dispersion exists in a sample statistic. When dealing with proportions, it gives insight into how your sample proportion varies from the actual population proportion.

The formula to compute the standard error for a sample proportion is:
  • \( SE = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } \)
Where "\( \hat{p} \)" is the sample proportion, and "\( n \)" is the sample size. In this example, the sample proportion of people who believe retesting should occur is 0.66, and the sample size is 1018. Plugging these into the formula, you get \( SE \approx 0.0148 \).

What does this number tell us? A smaller SE would imply our estimate of the proportion is very close to the true population proportion, while a larger SE would suggest a higher level of variability. Hence, understanding and calculating the SE is critical to determining the margin of error and creating an accurate confidence interval.
Proportion
A proportion in statistics refers to a part or fraction of the total that carries a certain characteristic or attribute. For example, if we say 66% of adults think elderly drivers should retake a road test, this is a sample proportion.

Determining the sample proportion involves converting the percentage in question into a decimal by dividing by 100. In the given exercise, 66% becomes \( \hat{p} = 0.66 \). This is used in the standard error and confidence interval calculations to draw insights about the population as a whole.

Proportions are significant because they help in making inferences about the entire population based on a sample. By considering the proportion of the sample and using statistical metrics like the standard error and margin of error, researchers can approximate the sentiments or characteristics of an entire group of people. This is why getting an accurate sample proportion is vital in analysis, ensuring that the results reflect the true population as closely as possible.

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Most popular questions from this chapter

Sleep deprived transportation workers. The National Sleep Foundation conducted a survey on the sleep habits of randomly sampled transportation workers and a control sample of non-transportation workers. The results of the survey are shown below. \begin{tabular}{lccccc} & \multicolumn{4}{c} { Transportation Professionals } \\ \cline { 3 - 6 } & & & Truck & Train & Bux/1axi/Limo \\ & Control & Pilots & Drivers & Operators & Drivers \\ \hline Less than 6 hours of sleep & 35 & 19 & 35 & 29 & 21 \\ 6 to 8 hours of sleep & 193 & 132 & 117 & 119 & 131 \\ More than 8 hours & 64 & 51 & 51 & 32 & 58 \\ \hline Total & 292 & 202 & 203 & 180 & 210 \end{tabular} Conduct a hypothesis test to evaluate if these data provide evidence of a difference between the proportions of truck drivers and non-transportation workers (the control group) who get less than 6 hours of sleep per day, i.e. are considered sleep deprived.

Gender and color preference. A 2001 study asked 1,924 male and 3,666 female undergraduate college students their favorite color. A \(95 \%\) confidence interval for the difference between the proportions of males and females whose favorite color is black ( \(p_{\text {male }}-p\) female \()\) was calculated to be (0.02,0.06) . Based on this information, determine if the following statements are true or false, and explain your reasoning for each statement you identify as false. 2 (a) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) lower to \(6 \%\) higher than the true proportion of females whose favorite color is black. (b) We are \(95 \%\) confident that the true proportion of males whose favorite color is black is \(2 \%\) to \(6 \%\) higher than the true proportion of females whose favorite color is black. (c) \(95 \%\) of random samples will produce \(95 \%\) confidence intervals that include the true difference between the population proportions of males and females whose favorite color is black. (d) We can conclude that there is a significant difference between the proportions of males and females whose favorite color is black and that the difference between the two sample proportions is too large to plausibly be due to chance. (e) The \(95 \%\) confidence interval for \(\left(p_{\text {female }}-p_{\text {male }}\right)\) cannot be calculated with only the information given in this exercise.

Young Americans, Part I. About \(77 \%\) of young adults think they can achieve the American dream. Determine if the following statements are true or false, and explain your reasoning. \(^{27}\) (a) The distribution of sample proportions of young Americans who think they can achieve the American dream in samples of size 20 is left skewed. (b) The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since \(n \geq 30\). (c) A random sample of 60 young Americans where \(85 \%\) think they can achieve the American dream would be considered unusual. (d) A random sample of 120 young Americans where \(85 \%\) think they can achieve the American dream would be considered unusual.

Browsing on the mobile device. A 2012 survey of 2,254 American adults indicates that \(17 \%\) of cell phone owners do their browsing on their phone rather than a computer or other device. (a) According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones. " Conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of \(38 \%\) (b) Interpret the p-value in this context. (c) Calculate a \(95 \%\) confidence interval for the proportion of Americans who access the internet on their cell phones, and interpret the interval in this context.

True or false, Part I. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) The chi-square distribution, just like the normal distribution, has two parameters, mean and standard deviation. (b) The chi-square distribution is always right skewed, regardless of the value of the degrees of freedom parameter. (c) The chi-square statistic is always positive. (d) As the degrees of freedom increases, the shape of the chi-square distribution becomes more skewed.
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