/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Study time A graduate teaching a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Study time A graduate teaching assistant for Introduction to Statistics (STA 2023) at the University of Florida collected data from students in one of her classes in spring 2007 to investigate whether study time per week (average number of hours) differed between students in the class who planned to go to graduate school and those who did not. The data were as follows: Graduate school: \(\quad 15,7,15,10,5,5,2,3,12,16,15,37\) 8,14,10,18,3,25,15,5,5 No graduate school: 6,8,15,6,5,14,10,10,12,5 Using software or a calculator, a. Find the sample mean and standard deviation for each group. Interpret. b. Find the standard error for the difference between the sample means. Interpret. c. Find a \(95 \%\) confidence interval comparing the population means. Interpret.

Short Answer

Expert verified
Students planning to go to graduate school study more on average. The 95% CI of [0.07, 5.47] suggests a significant difference in mean study times.

Step by step solution

01

Calculate Sample Means

To find the sample mean for each group, sum up all the study time hours within that group and divide by the number of observations in the group. For graduate school: \[ \text{Mean}_1 = \frac{15 + 7 + 15 + 10 + 5 + 5 + 2 + 3 + 12 + 16 + 15 + 37 + 8 + 14 + 10 + 18 + 3 + 25 + 15 + 5 + 5}{21} = 11.76 \]For no graduate school: \[ \text{Mean}_2 = \frac{6 + 8 + 15 + 6 + 5 + 14 + 10 + 10 + 12 + 5}{10} = 9.1 \]
02

Calculate Sample Standard Deviations

Standard deviation measures the spread of the data around the mean. Use the formula for the sample standard deviation:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \]Perform this calculation for both groups to obtain:For graduate school: \[ s_1 = \approx 8.34 \]For no graduate school:\[ s_2 = \approx 3.86 \]
03

Calculate Standard Error of the Mean Difference

The standard error of the difference between two sample means is calculated using:\[ SE_{\text{difference}} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Substitute the standard deviations and sample sizes:\[ SE_{\text{difference}} = \sqrt{\frac{8.34^2}{21} + \frac{3.86^2}{10}} = \approx 2.56 \]
04

Calculate the 95% Confidence Interval for the Mean Difference

A 95% confidence interval for the difference in means can be calculated as:\[ \text{CI} = (\bar{x}_1 - \bar{x}_2) \pm t^* \times SE_{\text{difference}} \]Where \(t^*\) is the t-distribution critical value for the combined degrees of freedom (approximately 29). Using \(t^* \approx 2.045\):\[ \text{CI} = (11.76 - 9.1) \pm 2.045 \times 2.56 = [0.07, 5.47] \]
05

Interpretation of Results

The sample means suggest that students planning on graduate school study more (mean of 11.76 hours) compared to those not planning on it (mean of 9.1 hours). The calculated 95% confidence interval indicates that the true difference in mean study times between the groups is somewhere between 0.07 and 5.47 hours. This interval does not contain zero, indicating a significant difference in average study time between the two groups.

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 91Ó°ÊÓ!

Key Concepts

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

Confidence Interval
A confidence interval provides a range within which we can expect the true population parameter to fall, with a given level of confidence. In this exercise, the confidence interval is used to compare the average study times between students planning to attend graduate school and those who are not.

When we talk about a 95% confidence interval, it means we are 95% confident that the true difference between the study times of these two groups lies within the calculated range. This range is [0.07, 5.47] for our data. Because this interval does not include zero, it strongly suggests that there is a real difference in study habits between the two groups.

Confidence intervals rely on the t-distribution, especially when dealing with smaller sample sizes or unknown population variances.
  • The t-distribution accounts for the additional uncertainty in estimating the standard deviation from the sample.
  • To calculate the interval, a critical value from the t-distribution, called \(t^*\), is used.
  • In this case, \(t^*\) is approximately 2.045, which corresponds to 29 degrees of freedom.
Sample Mean
The sample mean is the average value of a sample, calculated by summing up all the observations in the sample and dividing by the number of observations. It is a key metric for statistical analysis, as it represents a central value around which data points are distributed.

In our analysis, we have calculated two sample means:
  • For students planning on going to graduate school, the sample mean study time was 11.76 hours per week.
  • For students not planning on graduate school, the sample mean was 9.1 hours per week.

The sample mean provides an estimate of the population mean, though it may not be exactly equal to the true population mean. However, when paired with measures of variability, like standard deviation, it becomes a powerful tool for comparison in hypothesis testing and interval estimation.
Standard Deviation
Standard deviation measures how spread out the numbers in a data set are around the mean. It's a crucial component of summary statistics and is particularly helpful to understand the variability within a sample.

For the groups in this exercise, the standard deviations were computed as follows:
  • Graduate school aspirants had a standard deviation of approximately 8.34 hours.
  • The group not aspiring for graduate school had a standard deviation of about 3.86 hours.

This tells us that study time among students aiming for graduate school is more varied than those who are not. Standard deviation is important because it reflects data consistency and can highlight potential differences in data groups.
Standard Error
Standard error gives an idea of how much variability exists between sample means if we were to take multiple samples. Specifically, it quantifies the precision of the sample mean as an estimate of the population mean.

In our example, the standard error for the difference in two group means calculates the variability between these two averages. It is derived from the standard deviations of each group and their sample sizes. The calculation in this problem provided a standard error of approximately 2.56.
  • The smaller the standard error, the more precise our estimate of the mean difference is.
  • It is crucial for creating confidence intervals and performing hypothesis tests.

The standard error formula incorporates both the size of each sample (larger samples provide more reliable estimates) and their respective variability, fortifying our understanding of the data's accuracy.

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

From the box formula for the standard error at the end of Section 10.1 \(\frac{s e(\text { estimate } 1-\text { estimate } 2)=}{\sqrt{[s e(\text { estimate } 1)]^{2}+[s e(\text { estimate } 2)]^{2}}}\) if you know the se for each of two independent estimates, you can find the se of their difference. This is useful because often articles report an se for each sample mean or proportion, but not the se or a confidence interval for their difference. Many medical studies have used a large sample of subjects from Framingham, Massachusetts, who have been followed since 1948. A study (Annual of Internal Medicine, vol. \(138,2003,\) pp. \(24-32\) ) estimated the number of years of life lost by being obese and a smoker. For females of age 40 , adjusting for other factors, the number of years of life left were estimated to have a mean of \(46.3(s e=0.6)\) for nonsmokers of normal weight and a mean of \(33.0(s e=1.8)\) for smokers who were obese. Construct a \(95 \%\) confidence interval for the population mean number of years lost. Interpret.

In Western Australia, handheld cell phone use while driving has been banned since \(2001,\) but hands-free devices are legal. A study (published in the British Medical Journal in 2005 ) of 456 drivers in Perth who had been in a crash observed if they were using a cell phone before the crash and if they were using a cell phone during an earlier period when no accident occurred. Thus, each driver served as his or her own control group in the study. a. In comparing rates of cell phone use before the crash and the earlier accident-free period, should we use methods for independent samples or for dependent samples? Explain. b. Identify a test you can use to see whether the proportion of drivers using a cell phone differs between the period before the crash and the earlier accident-free period.

The table shows results from the 2014 General Social Survey on gender and whether one believes in an afterlife. $$ \begin{array}{lccc} \hline & \ {\text { Belief in Afterlife }} & \\ { 2 - 3 } \text { Gender } & \text { Yes } & \text { No } & \text { Total } \\\ \hline \text { Female } & 1026 & 207 & 1233 \\ \text { Male } & 757 & 252 & 1009 \\ \hline\end{array}$$ a. Denote the population proportion who believe in an afterlife by \(p_{1}\) for females and by \(p_{2}\) for males. Estimate \(p_{1}, p_{2},\) and \(\left(p_{1}-p_{2}\right)\) b. Find the standard error for the estimate of \(\left(p_{1}-p_{2}\right)\). Interpret. c. Construct a \(95 \%\) confidence interval for \(\left(p_{1}-p_{2}\right)\). Can you conclude which of \(p_{1}\) and \(p_{2}\) is larger? Explain. d. Suppose that, unknown to us, \(p_{1}=0.81\) and \(p_{2}=0.72 .\) Does the confidence interval in part c contain the parameter it is designed to estimate? Explain.

A researcher in the College of Nursing, University of Florida, hypothesized that women who undergo breast augmentation surgery would gain an increase in self-esteem. The article about the study \(^{15}\) indicated that for the 84 subjects who volunteered for the study, the scores on the Rosenberg Self- Esteem Scale were 20.7 before the surgery (std. dev. \(=6.3\) ) and 24.9 after the surgery (std. \(\mathrm{dev}=4.6\) ). The author reported that a paired difference significance test had \(t=9.8\) and a P-value below 0.0001 . a. Were the samples compared dependent samples, or independent samples? Explain. b. Can you obtain the stated \(t\) statistic from the values reported for the means, standard deviation, and sample size? Why or why not?

A Time Magazine article titled "Wal-Mart's Gender Gap" (July 5,2004\()\) stated that in 2001 women managers at Wal-Mart earned \(\$ 14,500\) less than their male counterparts. a. If these data are based on a random sample of managers at Wal-Mart, what more would you need to know about the sample to determine whether this is a "statistically significant" difference? b. If these data referred to all the managers at Wal-Mart and if you can get the information specified in part a, is it relevant to conduct a significance test? Explain.

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.