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A 2017 Gallup poll reported that 658 out of 1028 U.S. adults believe that marijuana should be legalized. When Gallup first polled U.S. adults about this subject in 1969 , only \(12 \%\) supported legalization. Assume the conditions for using the CLT are met. a. Find and interpret a \(99 \%\) confidence interval for the proportion of U.S. adults in 2017 that believe marijuana should be legalized. b. Find and interpret a \(95 \%\) confidence interval for this population parameter. c. Find the margin of error for each of the confidence intervals found in parts a and \(\mathrm{b}\). d. Without computing it, how would the margin of error of a \(90 \%\) confidence interval compare with the margin of error for the \(95 \%\) and \(99 \%\) intervals? Construct the \(90 \%\) confidence interval to see if your prediction was correct.

Short Answer

Expert verified
a. The 99% confidence interval for the proportion of U.S. adults in 2017 that believe marijuana should be legalized is (\(L_1, U_1\)), where \(L_1\) and \(U_1\) are the lower and upper limits of the interval that you calculate. b. The 95% confidence interval for this population parameter is (\(L_2, U_2\)), where \(L_2\) and \(U_2\) are the lower and upper limits of the interval that you calculate. c. The margins of error for the 99% and 95% confidence intervals are \(E_1\) and \(E_2\) respectively, which you calculate. d. The margin of error of a 90% confidence interval would be less than the margins of error for the 95% and 99% confidence intervals. The 90% confidence interval that you construct is (\(L_3, U_3\)), where \(L_3\) and \(U_3\) are the lower and upper limits of the interval.

Step by step solution

01

Compute the sample proportion and critical values

Firstly, compute the sample proportion \(\hat{p}\) by dividing the number of successes (in this case, the number of adults who believe marijuana should be legalized) by the total number of trials. Then, find the critical values \(Z_{0.005}\), \(Z_{0.025}\) and \(Z_{0.05}\) from a standard normal distribution table.
02

Compute the 99% confidence interval

Insert the values of \(\hat{p}\), \(n\) and \(Z_{0.005}\) in the formula for the confidence interval to get the 99% confidence interval. The 99% confidence interval is calculated as \(\hat{p} ± Z_{0.005}\sqrt{\hat{p}(1 - \hat{p}) / n}\)
03

Compute the 95% confidence interval

Similarly, compute the 95% confidence interval by inserting the values of \(\hat{p}\), \(n\) and \(Z_{0.025}\) in the formula for the confidence interval. The 95% confidence interval is given as \(\hat{p} ± Z_{0.025}\sqrt{\hat{p}(1 - \hat{p}) / n}\)
04

Compute the margins of error for the confidence intervals

Use the formula for the margin of error to compute the margins of error for the 99% and 95% confidence intervals. The margin of error is calculated as \(Z\sqrt{\hat{p}(1 - \hat{p}) / n}\)
05

Compare and construct the 90% confidence interval

The margin of error of a 90% confidence interval would be less than the margins of error for the 95% and 99% confidence intervals because as the level of confidence decreases, the margin of error decreases. Then, compute the 90% confidence interval by inserting the values of \(\hat{p}\), \(n\) and \(Z_{0.05}\) in the formula for the confidence interval. The 90% confidence interval is given as \(\hat{p} ± Z_{0.05}\sqrt{\hat{p}(1 - \hat{p}) / n}\)

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Key Concepts

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

Margin of Error
The margin of error is a crucial component in understanding confidence intervals. It tells us how much we can expect our sample results to differ from the true population parameter. Essentially, it represents the range of values above and below the sample statistic within which we expect the population parameter to lie, based on a certain level of confidence.

This margin is calculated using the formula: \[ \text{Margin of Error} = Z \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \]where:
  • \(Z\) is the critical value that corresponds to a given level of confidence.
  • \(\hat{p}\) is the sample proportion.
  • \(n\) is the sample size.
A larger margin of error means more variability in the estimate, meaning the confidence interval is wider. Conversely, a smaller margin of error signifies a more precise estimate. The margin of error declines as the level of confidence decreases and the sample size increases.
Sample Proportion
A sample proportion \(\hat{p}\) is an estimate of the population proportion, which indicates the fraction of individuals in a sample with a particular characteristic. To find the sample proportion, we divide the number of successful outcomes by the total sample size. For example, if 658 out of 1028 U.S. adults believe marijuana should be legalized, the sample proportion \(\hat{p}\) is calculated as follows:

\[ \hat{p} = \frac{658}{1028} \approx 0.64 \]This proportion serves as a point estimate for constructing confidence intervals and calculating the margin of error. Since it's derived from a sample, it might not perfectly reflect the entire population, so we use confidence intervals to provide a range of plausible values for the population proportion.
Critical Values
Critical values are key in calculating confidence intervals and are derived from the standard normal distribution. They define the cut-off points that determine the level of confidence we have in our interval estimate. The critical value \(Z\) changes based on the desired level of confidence.
  • The 99% confidence level uses critical value \(Z_{0.005}\).
  • The 95% confidence level uses critical value \(Z_{0.025}\).
  • The 90% confidence level uses critical value \(Z_{0.05}\).
The \(Z\) values are found using statistical tables or calculators that give the \(Z\)-score corresponding to the tails of a standard normal distribution. Higher confidence levels require larger critical values, which lead to wider confidence intervals, capturing a broader range of possible values for the population parameter.
Central Limit Theorem (CLT)
The Central Limit Theorem (CLT) is a fundamental principle in statistics that simplifies the process of making inferences about populations. It states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the population's distribution. This is remarkable because it allows us to use normal distribution properties even when dealing with non-normal populations.

For proportion estimates, like in our Gallup poll example, the CLT suggests that if the sample size is large enough, the distribution of the sample proportion \(\hat{p}\) will also be approximately normal. This approximation allows us to construct confidence intervals using normal distribution critical values. Often, the rule of thumb is that both \(np\) and \(n(1-p)\) should be at least 10 for the CLT to apply effectively, ensuring the sample proportion's distribution is normal.

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Most popular questions from this chapter

The 2017 Chapman University Survey of American Fears asked a random sample of 1207 adults Americans if they believed that aliens had come to Earth in modern times, and \(26 \%\) responded yes. a. What is the standard error for this estimate of the percentage of all Americans who believe that aliens have come to Earth in modern times? b. Find a \(95 \%\) confidence interval for the proportion of all Americans who believe that aliens have come to Earth in modern times. c. What is the margin of error for the \(95 \%\) confidence interval? d. A similar poll conducted in 2016 found that \(24.7 \%\) of Americans believed aliens have come to Earth in modern times. Based on your confidence interval, can you conclude that the proportion of Americans who believe this has increased since \(2016 ?\)

Suppose it is known that \(20 \%\) of students at a certain college participate in a textbook recycling program each semester. a. If a random sample of 50 students is selected, do we expect that exactly \(20 \%\) of the sample participates in the textbook recycling program? Why or why not? b. Suppose we take a sample of 500 students and find the sample proportion participating in the recycling program. Which sample proportion do you think is more likely to be closer to \(20 \%\) : the proportion from a sample size of 50 or the proportion from a sample size of \(500 ?\) Explain your reasoning.

The website www.mlb.com compiles statistics on all professional baseball players. For the 2017 season, statistics were recorded for all 663 players. Of this population, the mean batting average was \(0.236\) with a standard deviation of \(0.064\). Would it be appropriate to use this data to construct a \(95 \%\) confidence interval for the mean batting average of professional baseball players for the 2017 season? If so, construct the interval. If not, explain why it would be inappropriate to do so.

In 2018 it was estimated that approximately \(45 \%\) of the American population watches the Super Bowl yearly. Suppose a sample of 120 Americans is randomly selected. After verifying the conditions for the Central Limit Theorem are met, find the probability that at the majority (more than \(50 \%\) ) watched the Super Bowl. (Source: vox.com).

Is simple random sampling usually done with or without replacement?

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