/*! 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 49 A study was performed among 40 b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A study was performed among 40 boys in a school in Edinburgh to look at the presence of spermatozoa in urine samples according to age \([15] .\) The boys entered the study at \(8-11\) years of age and left the study at \(12-18\) years of age. A 24-hour urine sample was supplied every 3 months by each boy. Table 10.28 gives the presence or absence of sperm cells in the urine samples for each boy together with the ages at entrance and exit of the study and the age at the first sperm-positive urine sample. For all parts of this question, exclude boys who exited this study without 1 sperm-positive urine sample (i.e., boys 8,9,14,25,28,29,30) Suppose mean age at spermatogenesis = 13.67 years, with standard deviation \(=0.89\) years and we assume that the age at spermatogenesis follows a normal distribution. The pediatrician would like to know what is the earliest age (in months) before which \(95 \%\) of boys experience spermatogenesis because he or she would like to refer boys who haven't experienced spermatogenesis by this age to a specialist for further follow-up. Can you estimate this age from the information provided in this part of the problem?

Short Answer

Expert verified
The earliest age before which 95% of boys experience spermatogenesis is 173 months.

Step by step solution

01

Understand the Normal Distribution

The problem states that the age at spermatogenesis follows a normal distribution with a mean of 13.67 years and a standard deviation of 0.89 years. We need to find the age where 95% of the boys have experienced spermatogenesis. This equates to identifying a specific percentile (the 95th percentile) of the normal distribution.
02

Identify the Z-Score for the 95th Percentile

For a normal distribution, the 95th percentile corresponds to a Z-score of approximately 1.645 (from Z-tables). This Z-score indicates how many standard deviations away from the mean the 95th percentile is.
03

Calculate the Corresponding Age

Use the formula for converting a Z-score to a data value: \[ X = ext{mean} + Z imes ext{standard deviation} \]Substitute the values to find the age:\[ X = 13.67 + 1.645 \times 0.89 \]Calculate the result.
04

Convert the Age from Years to Months

First, compute the age in years from the expression in Step 3. Then, convert the age from years to months by multiplying the age by 12, since there are 12 months in a year. This will give us the age in months.

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.

Normal Distribution
In the study of age at spermatogenesis, a normal distribution is used to model the ages at which boys first show signs of sperm production. The normal distribution is a fundamental concept in statistics characterized by its symmetrical, bell-shaped curve. It is determined by two critical parameters: the mean and the standard deviation. The mean is the arithmetic average of all ages, which in our study is 13.67 years. It represents the center of our distribution. The standard deviation, here being 0.89 years, gives us an idea of the spread of our data points around the mean. One of the key properties of a normal distribution is that it allows us to make probabilistic statements about the data. For this study, we are interested in the 95th percentile, meaning the age before which 95% of the boys are expected to have experienced spermatogenesis. The 95th percentile is found a specific number of standard deviations- typically represented by a Z-score - above the mean.
Pediatric Study
A pediatric study typically deals with health data concerning children and adolescents. In this study, researchers collected data on the presence of spermatozoa among boys aged between 8-18 years. Every three months, urine samples were collected to monitor when each boy first showed signs of spermatogenesis. Pediatric studies like these are crucial for understanding developmental milestones and guiding health interventions. The goal of this study was to determine an appropriate reference age by which boys should meet this developmental milestone. Those who do not may require further health evaluations as a precautionary measure. The study's ethical considerations include ensuring the privacy and confidentiality of the participants' data, particularly given the sensitivity of adolescent health records.
Statistical Analysis
To determine the age at which 95% of the boys in this study first experience spermatogenesis, a statistical analysis was conducted using the data collected. This involves calculating percentiles in the context of a normal distribution. A key step is finding the Z-score that corresponds to the 95th percentile.The Z-score for this percentile is approximately 1.645. Once the Z-score is identified, it is used in conjunction with the mean and standard deviation to calculate the corresponding age. The formula to convert a Z-score to an age is: \[ X = \text{mean} + Z \times \text{standard deviation} \]Substituting the given mean and standard deviation yields the age of interest. Such calculations allow researchers to make informed decisions about which participants might need additional follow-up.
Age at Spermatogenesis
Spermatogenesis refers to the process by which male germ cells develop into mature spermatozoa. The age at which boys undergo spermatogenesis can vary but often signals the onset of puberty. In this study, researchers aimed to identify the age by which a majority of boys reach this milestone. The study reveals a mean age of 13.67 years for spermatogenesis, with the calculation for the 95th percentile, or the earliest age before which 95% have experienced it, being critical for clinical decision-making. By using a normal distribution model, it was possible to determine this specific age in months, providing useful information for pediatricians to decide when a referral to a specialist might be warranted. Understanding this age range helps healthcare professionals monitor adolescent development and detect any anomalies early on.

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 study was performed among patients with progressive chronic kidney disease to assess the optimal time for the initiation of dialysis (Cooper et al. [22]). Patients were randomized to either i. an early start group, where dialysis was initiated when their estimated glomerular filtration rate (GFR) was between 10.0 and \(15.0 \mathrm{mL}\) per minute per \(1.73 \mathrm{m}^{2}\) of body surface area or ii. a late start group, where dialysis was initiated when their GFR was 5.0 to \(7.0 \mathrm{mL}\) per minute per \(1.73 \mathrm{m}^{2}\) of body surface area.The primary outcome was total mortality. There were 404 patients assigned to the early start group, of whom 152 died. There were 424 patients assigned to the late start group, of whom 155 died over the same time period. Patients were enrolled from July 2000 to November 2008 and were followed for 3 years. What test can be used to compare the mortality rates in the two groups?

A study group of 576 working women \(30-49\) years of age who took phenacetin- containing analgesics and a control group of 533 comparably aged women without such intake were identified in 1968 and followed for mortality and morbidity outcomes. One hypothesis to be tested was that phenacetin intake may influence renal (kidney) function and hence have an effect on specific indices of renal morbidity and mortality. The mortality status of these women was determined from 1968 to \(1987 .\) The researchers found that 16 of the women in the study group and 1 of the women in the control group died, where at least one cause of death was considered renal [13]. What statistical test should be used to compare the total mortality experience of the study group with that of the control group?

A 1980 study investigated the relationship between the use of OCs and the development of endometrial cancer [9]. The researchers found that of 117 endometrial-cancer patients, 6 had used the OC Oracon at some time in their Iives, whereas 8 of the 395 controls had used this agent. Test for an association between the use of Oracon and the incidence of endometrial cancer, using a two-tailed test.

It is fairly well known that perception of weight by adolescents does not always agree with actual weight. What is less clear is whether perception of weight differs by gender. For this purpose, a study was performed among students in a local high school, where students provided their actual height and weight by self-report. The following data were obtained from 286 students \((143 \text { boys and } 143\) girls). (The data for this problem were provided by Laura Rosner.) The students were classified as underweight if their bodymass index (BMI) \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\) was less than \(18.0 \mathrm{kg} / \mathrm{m}^{2}\), as normal if their BMI was \(\geq 18.0\) and \(<25.0,\) and overweight if their BMI was \(\geq 25.0\). Based on these criteria, 17 of the girls were underweight, 113 were of normal weight, and 13 were overweight. For the boys, 7 were underweight, 115 were of normal weight, and 21 were overweight. What test procedure can be used to assess whether the weight status of boys significantly differs from girls?

Dry eye is the most prevalent form of ocular discomfort and irritation, with approximately 20 million people in the United States having mild to moderate dry eye. A small clinical trial was performed to compare the effectiveness of an active drug vs. placebo for relieving symptoms of dry eye. Specifically, patients were randomized to either active drug or placebo for 2 weeks. They then came for a clinic visit where they were exposed to a chamber with a controlled adverse environment (CAE) for 90 minutes (with low humidity intended to exacerbate symptoms of dry eye). The patients were then asked to report their degree of discomfort while in the CAE using the following scale: \((0=\text { none, } 1=\) intermittent awareness, \(2=\) constant awareness, \(3=\) intermittent discomfort, \(4=\) constant discomfort). The results by treatment group are shown in Table 10.45 What is the difference between a nominal and ordinal categorical variable? What type of variable is ocular discomfort?

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.