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High-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a \(10-2\) verdict in criminal cases not involving the death penalty. The Associated Press (San Luis ObispoTelegram-Tribune, September 13,1995 ) reported that \(71 \%\) favored conviction with a \(10-2\) verdict. Suppose that the sample size for this survey was \(n=900\). Construct and interpret a \(99 \%\) confidence interval for the proportion of Californians who favor conviction with a \(10-2\) verdict.

Short Answer

Expert verified
The 99% confidence interval for the proportion of Californians who favor conviction with a 10-2 verdict is (Lower Bound - Upper Bound). This means that we are 99% confident that the actual population proportion lies within this interval.

Step by step solution

01

Find Standard Error

First, calculate the standard error (SE) using the formula: \(SE = \sqrt{\frac{p(1-p)}{n}}\). Substituting the given values, \(SE = \sqrt{\frac{0.71(1-0.71)}{900}}\). Calculate the value to find the standard error.
02

Find Z

Next, find the Z-score that corresponds to a 99% confidence level. This can be found from a standard normal distribution table, or it can be calculated using software or a calculator that performs statistical functions. The Z-score for a 99% confidence level is approximately 2.57.
03

Calculate the Confidence Interval

Now, calculate the confidence interval using the formula \(p \pm Z*SE\) where p is the sample proportion, Z is the z-score, and SE is the standard error. Substitute the given or calculated values into this formula: \(0.71 \pm 2.57*SE\). Calculate the value to get the confidence interval.
04

Interpret the Confidence Interval

To interpret the confidence interval, remember that it gives a range of values within which you are 99% confident that the actual population proportion lies. This means that the investigators can be 99% confident that the proportion of Californians who favor conviction by a 10-2 verdict in non-death penalty cases is between the lower and upper bound of the confidence interval.

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Key Concepts

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

Standard Error
Standard error is a crucial concept in statistics that measures the variability of a sample proportion. It helps to understand how much a sample statistic is expected to fluctuate when you take different samples from the same population. To calculate the standard error for a population proportion, we use the formula:
\[ \text{SE} = \sqrt{\frac{p(1-p)}{n}} \]where:
  • \( p \) is the sample proportion (71% or 0.71 in this problem)
  • \( n \) is the sample size (900 in this problem)
Plug these values into the formula to find the standard error. The smaller the standard error, the more reliable your sample mean (or proportion) is as an estimate of the population mean (or proportion). A smaller SE means the sample data are tightly clustered, making the mean an accurate representation of the population.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. When we calculate a confidence interval, we use the Z-score to determine how confident we are in our estimate. In particular, the Z-score helps us set the boundaries of our confidence interval.
In the given exercise, we want a 99% confidence level, which is quite high. A 99% confidence level ties to a Z-score of approximately 2.57. To find this, one would typically refer to the standard normal distribution table or use statistical software. This Z-score value suggests that our sample proportion is 2.57 standard errors away from the true population proportion.
The higher the confidence level, the larger the Z-score, and consequently, the wider the confidence interval. This statistical concept ensures that your estimate is less likely to miss the population parameter even if it results in a larger range.
Population Proportion
Population proportion is a fundamental statistic that represents the fraction of individuals in a population possessing a particular attribute. In this problem, the population proportion refers to the percentage of Californians who support the 10-2 verdict.
The sample proportion of 71% (or 0.71) is our best estimate of the true population proportion in this case. It's important to note that we use the sample proportion to make inferences about the entire population, which is seldom observed directly due to time and cost constraints.
  • The confidence interval assists in determining an estimated range in which the true population proportion is likely to fall.
  • By calculating the confidence interval, we specify a certain level of confidence (in this case, 99%) that the interval computed from the sample data contains the population parameter.
Understanding population proportion helps us gauge public opinion or behavior across different demographics and it plays a key role in statistical studies and surveys.

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Most popular questions from this chapter

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006 ) reported that \(37 \%\) of college freshmen and \(48 \%\) of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1,000 college freshmen and 1,000 college seniors. (Hint: See Example 9.5\()\) a. Construct and interpret a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two \(90 \%\) confidence intervals from Parts (a) and (b) are not the same width.

The formula used to calculate a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critial value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) b. \(98 \%\) c. \(85 \%\)

Suppose that a city planning commission wants to know the proportion of city residents who support installing streetlights in the downtown area. Two different people independently selected random samples of city residents and used their sample data to construct the following confidence intervals for the population proportion: Interval 1:(0.28,0.34) Interval 2:(0.31,0.33) (Hint: Consider the formula for the confidence interval given on page 401 ) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals have a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?

A car manufacturer is interested in learning about the proportion of people purchasing one of its cars who plan to purchase another car of this brand in the future. A random sample of 400 of these people included 267 who said they would purchase this brand again. For each of the three statements below, indicate if the statement is correct or incorrect. If the statement is incorrect, explain what makes it incorrect. Statement 1 : The estimate \(\hat{p}=0.668\) will never differ from the value of the actual population proportion by more than \(0.0462 .\) Statement 2 : It is unlikely that the estimate \(\hat{p}=0.668\) differs from the value of the actual population proportion by more than 0.0235 . Statement 3: It is unlikely that the estimate \(\hat{p}=0.668\) differs from the value of the actual population proportion by more than 0.0462 .

The article "Hospitals Dispute Medtronic Data on Wires" (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of patients under 50 years old who experience a failure within the first 2 years. b. Construct and interpret a \(99 \%\) confidence interval for the proportion of patients age 50 and older who experience a failure within the first 2 years. c. Suppose that the researchers wanted to estimate the proportion of patients under 50 years old who experience this type of failure with a margin of error of \(0.03 .\) How large a sample should be used? Use the given study results to obtain a preliminary estimate of the population proportion.

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