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How Much More Effective Is It to Test Your- self in Studying? In Exercise \(5.23,\) we see that students who study by giving themselves quizzes recall study by reading. In Exercise 5.23 we see that there is an effect, but often the question of interest is not "Is there an effect?" but instead "How big is the effect?" To address this second question, use the fact that \(\hat{p}_{Q}=0.42\) and \(\hat{p}_{R}=0.15\) to find and interpret a \(99 \%\) confidence interval for the difference in proportions \(p_{Q}-p_{R},\) where \(p_{Q}\) represents the proportion of items correctly recalled by all students who study using a self-quiz method and \(p_{R}\) represents the proportion of items correctly recalled by all students who study using a reading-only approach. Assume that the standard error for a bootstrap distribution of such differences is about 0.07

Short Answer

Expert verified
The 99% confidence interval for the difference in proportions between the two studying methods is approximately (0.0901, 0.4499). This suggests that, with 99% confidence, the true difference in recall effectiveness between self-quizzing and reading only can range from 0.0901 to 0.4499 in favor of self-quizzing. This implies that self-quizzing is likely to be more effective than reading only.

Step by step solution

01

Determine the Point Estimate

The point estimate for the difference in proportions between the two studying methods is given by the formula \(\hat{p}_{Q} - \hat{p}_{R}\), which is 0.42 - 0.15 which equals 0.27.
02

Determine Margin of Error

The margin of error at 99% confidence level (for z distribution it is 2.57). This is the product of the z-score and standard error. The calculation can be achieved with 2.57 * 0.07 which equals 0.1799.
03

Calculate the Confidence Interval

The 99% confidence interval can be computed by subtracting and adding the margin of error from the point estimate. Lower bound is 0.27 - 0.1799 = 0.0901, and upper bound is 0.27 + 0.1799 = 0.4499. Thus, the confidence interval is (0.0901, 0.4499).

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Key Concepts

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

Difference in Proportions
When comparing two different populations or groups, one essential statistical measurement is the difference in proportions. This involves computing the difference between the proportion of successes or particular outcomes in one group (\( p_Q \)) and the proportion in another group (\( p_R \)). For instance, let's say we're looking to compare two study methods. One method is a self-quiz approach, and the other is a reading-only strategy.

By calculating the difference between the proportion of students who recall information after self-quizzing (\( \text{hat}{p}_Q \text{=} 0.42 \text{(42%)} \text{recall rate} \text{for self-quizzing} \)) and those who recall after reading-only (\( \text{hat}{p}_R \text{=} 0.15 \text{(15%)} \text{recall rate} \text{for reading-only} \)), we can measure the effectiveness of each study method. In this scenario, the point estimate of the difference is 0.27 or 27%, indicating that the self-quiz method, on average, yields a 27% higher recall rate than the reading-only approach.

Understanding the difference in proportions deepens our comprehension of how two methods compare and helps to measure the effectiveness of one relative to the other. In educational settings, such insights can be crucial for tailoring more effective study techniques.
Bootstrap Distribution
The bootstrap distribution is a modern statistical technique used to estimate the sampling distribution of a statistic by resampling with replacement from the original data. It simulates the process of taking many samples from a population and creates a large number of 'resampled' datasets. For example, when evaluating the difference in proportions mentioned earlier, we may not have the entire population's data, but we can use a bootstrap distribution to estimate it.

In our case involving study methods, the standard error of 0.07 is presumed to be from a bootstrap distribution of the difference in proportions. This value quantifies the variability of the estimated differences that we would observe if we repeated our study method comparison multiple times, resampling from the student populations each time. The lower the standard error, the less variability there is in the estimated differences, indicating more reliable results.

Bootstrap methods are exceptionally valuable when the sample size is too small or when the theoretical distribution of a statistic is complex or unknown. This flexibility makes it an important tool in modern statistics, particularly in fields like psychological research or educational sciences where repeated sampling can be challenging.
Margin of Error
The margin of error plays a critical role in understanding the range within which the true population parameter is likely to lie. It defines the bounds of the confidence interval, accounting for the inherent uncertainty in sampling. The margin of error is affected by both the level of confidence we desire (for example, 99%) and the standard error.

In our study methods example, to determine the margin of error at a 99% confidence level, we use the standard error (from the bootstrap distribution) in conjunction with the z-score that corresponds to the desired confidence level. Here, a z-score of 2.57 reflects the 99% confidence level, and the standard error is 0.07. Multiplying them gives us a margin of error of approximately 0.1799, or about 18%.

This margin of error is then applied to the point estimate of the difference in proportions, in both directions, to establish a confidence interval. The result is an interval within which we are 99% confident the true difference in recall rates lies if the study were to be repeated many times. It is crucial for interpreting the precision of the estimated difference in proportions and making informed decisions in educational practice or policy.

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Most popular questions from this chapter

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a mean \(\mu\) if the sample has \(n=50\) with \(\bar{x}=72\) and \(s=12,\) and the standard error is \(S E=1.70 .\)

5.55 Correlation between Time and Distance in Commuting In Exercise B.62 on page \(363,\) we find an interval estimate for the correlation between Distance (in miles) and Time (in minutes) for Atlanta commuters, based on the sample of size \(n=\) 500 in CommuteAtlanta. The correlation in the original sample is \(r=0.807\). (a) Use technology and a bootstrap distribution to estimate the standard error of sample correlations between Distance and Time for samples of 500 Atlanta commutes. (b) Assuming that the bootstrap correlations can be modeled with a normal distribution, use the results of (a) to find and interpret a \(90 \%\) confi\(\mathrm{s}\) dence interval for the correlation between dis- \({ }^{23} \mathrm{ht}\) tance and time of Atlanta commutes.

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu=10\) vs \(H_{a}: \mu \neq 10\) when the sample has \(n=75, \bar{x}=11.3,\) and \(s=0.85,\) with \(S E=0.10\).

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) when the sample has \(n=50\) and \(\hat{p}=0.41,\) with \(S E=0.07\).

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