Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

You want to estimate the difference in grade point averages between two groups of college students accurate to within .2 grade point, with probability approximately equal to \(.95 .\) If the standard deviation of the grade point measurements is approximately equal to .6, how many students must be included in each group? (Assume that the groups will be of equal size.)

Short Answer

Expert verified
Answer: To estimate the difference in grade point averages between the two groups with the given accuracy and probability, we need to have at least 35 students in each group.

Step by step solution

01

Find the Z-score for the desired confidence level

To find the Z-score for a 95% confidence level, we can consult a Z-table or use a calculator that can compute it for us. We are actually looking for the Z-score that corresponds to cumulative probability 0.975 (since we want the probability to be symmetric around the mean). By looking up a Z-table or using a calculator, we find that the Z-score for a 95% confidence level is approximately 1.96.
02

Plug the values into the sample size formula

Now that we have all the necessary information, we can plug the values into the formula for the sample size estimation: \(n = \left(\frac{1.96 \cdot 0.6}{0.2}\right)^2\)
03

Calculate the sample size

Let's calculate the sample size: \(n = \left(\frac{1.96 \cdot 0.6}{0.2}\right)^2 = \left(\frac{1.176}{0.2}\right)^2 \approx 34.57\) Since we cannot have a fraction of a student, we need to round the number up to the nearest whole number: \(n = 35\)
04

Interpret the result

So, to estimate the difference in grade point averages between the two groups of college students accurate to within 0.2 grade point and with a probability of approximately 0.95, we need to have at least 35 students in each group.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Confidence Level
When estimating a statistical measure, such as the difference in grade point averages, we must decide how confident we want to be in our estimate. This is where the confidence level comes into play. It is a percentage that reflects how sure we are that the true parameter falls within the estimated range. So, for a 95% confidence level, we're saying there's a 95 out of 100 chance that the true difference in grade point averages falls within our calculated interval. High confidence levels are associated with a broader interval, meaning we might need a larger sample size, but we can be more assured of our results.

To achieve the desired confidence level, we must use an appropriate Z-score as part of our calculations. For a 95% confidence level, the Z-score indicates the number of standard deviations a point is from the mean of a distribution, encompassing 95% of the data in a normal distribution. It's important to understand that the higher the confidence level, the larger the Z-score and the wider the confidence interval. This ensures that we are accounting for more variation and uncertainty in our estimate.
Z-score
The Z-score, also known as a standard score, quantifies how many standard deviations an element is from the mean of a distribution. In the context of sample size estimation, it's used as a multiplier to ensure that our confidence interval is appropriately wide to include the true mean with the desired level of confidence. In simpler terms, the Z-score is a way of mapping the distance of a score (or the difference in scores) to a number that represents our certainty in the context of a normal distribution.

For a 95% confidence level, the Z-score is approximately 1.96. This means that to be 95% confident, the range where we believe the true mean lies should extend 1.96 standard deviations from the sample mean on both sides. This is central to calculating the sample size needed because, as the required confidence level increases, so does the corresponding Z-score. As this value grows, we need more data to confidently say a certain interval contains the true mean.
Standard Deviation
Understanding standard deviation is crucial when talking about sample sizes and confidence intervals. Standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In the context of our example, the standard deviation of grade point averages is 0.6. This number is used to calculate the range around the estimated mean where we expect our true population mean to lie. It impacts the width of the confidence interval: the greater the standard deviation, the wider the interval for a given confidence level. Therefore, knowledge of the standard deviation helps in determining how large our sample size should be to achieve a certain precision in our estimates.
Grade Point Average
In educational contexts, the grade point average (GPA) is a standard way of measuring academic achievement. It is often calculated as the average of grades achieved in courses, weighted by the credit hours or units of each course. GPAs are commonly used for comparing the academic performances of individuals or groups.

In our exercise, we are examining the difference between the GPAs of two groups of students. The goal is to estimate this difference with a certain level of precision and confidence. Accurate to within 0.2 grade points means that our estimate of the difference will not vary more than 0.2 from the true difference. In statistical terms, the margin of error, defined based on our sample size and variability (i.e., standard deviation), will dictate how precise our estimate can be. When the desired precision is high, we usually need a larger sample size to ensure our confidence interval for GPA difference is sufficiently small.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A USA Today snapshot reported the results of a random sample of 500 women who were asked what reasons they might have to consider dieting, \({ }^{5}\) with the following results. a. Find a \(95 \%\) confidence interval for the proportion of all women who would consider dieting to improve their health. b. Find a \(90 \%\) confidence interval for the proportion of all women who would consider dieting in order to have more energy.

Suppose you want to estimate one of four parameters- \(\mu, \mu_{1}-\mu_{2}, p,\) or \(p_{1}-p_{2}-\) to within a given bound with a certain amount of confidence. Use the information given to find the appropriate sample size(s). Estimating \(p\) to within .04 with probability .95. You suspect that \(p\) is equal to some value between .1 and \(.3 .\)

Baseball Fans The first day of baseball comes in late March, ending in October with the World Series. Does fan support grow as the season goes on? Two CNN/USA Today/Gallup polls, one conducted in March and one in November, both involved random samples of 1001 adults aged 18 and older. In the March sample, \(45 \%\) of the adults claimed to be fans of professional baseball, while \(51 \%\) of the adults in the November sample claimed to be fans. \({ }^{19}\) a. Construct a \(99 \%\) confidence interval for the difference in the proportion of adults who claim to be fans in March versus November. b. Does the data indicate that the proportion of adults who claim to be fans increases in November, around the time of the World Series? Explain.

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given in Exercises \(1-2 .\) Construct a \(95 \%\) and a \(99 \%\) confidence interval for the difference in the population proportions. What does the phrase "95\% confident" or "99\% confident" mean? $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 500 & 500 \\\\\text { Number of Successes } & 120 & 147 \\\\\hline\end{array}$$

A survey is designed to estimate the proportion of sports utility vehicles (SUVs) being driven in the state of California. A random sample of 500 registrations are selected from a Department of Motor Vehicles database, and 68 are classified as SUVs. a. Use a \(95 \%\) confidence interval to estimate the proportion of SUVs in California. b. How can you estimate the proportion of SUVs in California with a higher degree of accuracy? (HINT: There are two answers.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free