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

Babies born extremely prematurely run the risk of various neurological problems and tend to have lower IQ and verbal ability scores than babies that are not premature. The article "Premature Babies May Recover Intelligence, Study Says" (San Luis Obispo Tribune, February 12,2003 ) summarized medical research that suggests that the deficits observed at an early age may decrease as children age. Children who were born prematurely were given a test of verbal ability at age 3 and again at age 8 . The test is scaled so that a score of 100 would be average for normal-birth-weight children. Data for 50 children who were born prematurely were used to generate the accompanying Minitab output, where Age 3 represents the verbal ability score at age 3 and Age8 represents the verbal ability score at age \(8 .\) Use the Minitab output to determine if there is convincing evidence that the mean verbal ability score for children born prematurely increases between age 3 and age 8 . You can assume that it is reasonable to regard the sample of 50 children as a random sample from the population of all children born prematurely. $$ \begin{aligned} &\text { Paired T-Test and Cl: Age8, Age3 }\\\ &\begin{array}{l} \text { Paired } T \text { for } \text { Age8 - Age3 } \\ \begin{array}{rrrrr} & \text { N } & \text { Mean } & \text { StDev } & \text { Se Mean } \\ \text { Age8 } & 50 & 97.21 & 16.97 & 2.40 \\ \text { Age3 } & 50 & 87.30 & 13.84 & 1.96 \\ \text { Difference } & 50 & 9.91 & 22.11 & 3.13 \\ \text { T-Test of mean difference } & =0(\mathrm{vs}>0): \text { T-Value }=3.17 \\ \text { P-Value }=0.001 & & & \end{array} \end{array} \end{aligned} $$

Short Answer

Expert verified
There is statistically significant evidence to suggest that children born prematurely experience an increase in verbal ability score between the ages 3 and 8, thus supporting the study's claim that deficits observed at an early age may decrease as children age.

Step by step solution

01

Understand the t-test report

In the given paired t-test report, the t-value is 3.17 and the p-value is 0.001. The t-test is used to determine whether the mean difference of two sets of data is significantly different from zero. P-value in a hypothesis test is used as a criterion for rejection or non-rejection of null hypothesis.
02

Formulate the null and alternative hypotheses

In this case, we test the null hypothesis \( H_0 : \mu_d = 0 \) (Mean difference in the scores is zero or they are same at both ages) against the alternative hypothesis \( H_a : \mu_d > 0 \) (Mean difference in the scores is greater than zero or the scores have improved).
03

Interpreting the t-test

The t-value of 3.17 means that the difference in means is 3.17 times as large as what would be expected by chance variation (the standard error of the difference). If p-value is less than the significance level (0.05), we reject the null hypothesis.
04

Checking the p-value to arrive at a conclusion

Given that the p-value is 0.001, which is less than the significance level of 0.05, we can reject the null hypothesis and infer that there is a significant increase in the verbal ability score from age 3 to age 8. Therefore, the study's claim is supported by this statistical analysis.

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.

Hypothesis Testing
Hypothesis testing is a fundamental concept in statistics used to determine if there's evidence in a sample of data to conclude that a certain condition holds for the entire population. It starts by assuming the null hypothesis (\( H_0 \)) is true, which in our scenario presupposes that the mean verbal ability score of premature babies at age 3 is equal to their score at age 8.

An alternative hypothesis (\( H_a \)or \( H_1 \) is formulated to represent a condition that contradicts the null hypothesis: for instance, it could state that the mean score at age 8 is higher than at age 3. Through the paired t-test, as in the provided exercise, we calculate a p-value that quantifies the probability of observing the data or something more extreme if the null hypothesis were true.

A low p-value (typically below 0.05) suggests that the observed data is unlikely under the null hypothesis, prompting analysts to reject \( H_0 \) in favor of \( H_a \). Hence, a p-value of 0.001 is strong evidence against the null hypothesis, pointing toward a real increase in verbal ability scores as these children aged.
Verbal Ability Score Analysis
Verbal ability score analysis involves examining and interpreting the scores from tests that measure verbal skills. In the exercise example, premature children's verbal ability scores were examined at age 3 and again at age 8. Here, we employ statistical techniques like the paired t-test to analyze these scores over time.

The paired t-test is particularly useful in this context because it looks at two related samples; the same group of children tested at different ages. It assesses whether their mean scores at one time point are significantly different from another. An important consideration is the mean difference, which is the average change in scores for the sample. In our case, a mean difference of 9.91 suggests an improvement. Coupled with the p-value, this analysis indicates whether the change in verbal ability scores is statistically significant.
Neurological Development in Premature Babies
Premature babies often face challenges that can affect their neurological development. This development is a complex process that can be influenced by a range of factors, including the level of care they receive and the environmental and genetic influences they're subject to post-birth.

Premature infants are at a higher risk of developmental delays, which might impact their cognitive abilities, including verbal skills. Standardized tests, such as the one employed in the exercise, can track these abilities and offer insights into their developmental trajectory.

Studies and analyses like the one referenced in the exercise are crucial for understanding how premature infants develop over time. They also serve as a basis for developing interventions aimed at mitigating the developmental deficits that could be a consequence of their early birth.

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

The Oregon Department of Health web site provides information on cost-to- charge ratio (the percentage of billed charges that are actual costs to the hospital). The following table gives cost-to-charge ratios for both inpatient and outpatient care in 2002 for a random sample of six hospitals in Oregon. $$ \begin{array}{ccc} & \begin{array}{c} 2002 \\ \text { Inpatient } \end{array} & \begin{array}{c} 2002 \\ \text { Hospital } \end{array} & \begin{array}{c} \text { Ratio } \\ \text { Ratient } \end{array} \\ \hline 1 & 68 & \text { Ratio } \\ 2 & 100 & 54 \\ 3 & 71 & 75 \\ 4 & 74 & 53 \\ 5 & 100 & 56 \\ 6 & 83 & 74 \\ & 88 \end{array} $$ Is there evidence that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care? Use a significance level of \(0.05 .\)

The authors of the paper "Ultrasound Techniques Applied to Body Fat Measurement in Male and Female Athletes" (Journal of Athletic Training [2009]: \(142-147\) ) compared two different methods for measuring body fat percentage. One method uses ultrasound and the other method uses X-ray technology. The accompanying table gives body fat percentages for 16 athletes using each of these methods (a subset of the data given in a graph that appeared in the paper). For purposes of this exercise, you can assume that the 16 athletes who participated in this study are representative of the population of athletes. Do these data provide convincing evidence that the mean body fat percentage measurement differs for the two methods? Test the appropriate hypotheses using \(\alpha=0.05\). $$ \begin{array}{crr} \text { Athlete } & \text { X-ray } & \text { Ultrasound } \\ \hline 1 & 5.00 & 4.75 \\ 2 & 7.00 & 3.75 \\ 3 & 9.25 & 9.00 \\ 4 & 12.00 & 11.75 \\ 5 & 17.25 & 17.00 \\ 6 & 29.50 & 27.50 \\ 7 & 5.50 & 6.50 \\ 8 & 6.00 & 6.75 \\ 9 & 8.00 & 8.75 \\ 10 & 8.50 & 9.50 \\ 11 & 9.25 & 9.50 \\ 12 & 11.00 & 12.00 \\ 13 & 12.00 & 12.25 \\ 14 & 14.00 & 15.50 \\ 15 & 17.00 & 18.00 \\ 16 & 18.00 & 18.25 \end{array} $$

In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: \(2024-2033\) ). The following statement appeared in the paper: "When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors \((\$ 313,205\) vs. \(\$ 521,560, P=0.004)\) a. What hypotheses did the researchers test to reach the stated conclusion? b. Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. \(\quad t=5.00\) iii. \(t=2.33\) ii. \(t=2.65\) iv. \(t=1.47\)

Do girls think they don't need to take as many science classes as boys? The article "Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences" (Science Education [1999]: \(55-76\) ) describes a survey of randomly selected children in grades \(4,5,\) and 6 . The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. a. Explain why these data are paired. b. The mean of the differences was -0.83 (indicating girls intended, on average, to take fewer science classes than they thought boys should take), and the standard deviation was 1.51 . Construct and interpret a \(95 \%\) confidence interval for the mean difference.

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work (University of Central Florida Undergraduate Research Journal, Spring 2005\()\) : $$ \begin{array}{cccc} & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Mean } \\ \text { GPA } \end{array} & \begin{array}{c} \text { Sandard } \\ \text { Deviation } \end{array} \\ \begin{array}{c} \text { Students Who Are } \\ \text { Employed } \end{array} & 184 & 3.12 & 0.485 \\ \begin{array}{c} \text { Students Who Are } \\ \text { Not Employed } \end{array} & 114 & 3.23 & 0.524 \\ \hline \end{array} $$ The samples were selected at random from working and nonworking students at the University of Central Florida. Does this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed?

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