The document discusses Chi Square distribution and analysis of frequency using Fisher's exact test and McNemar's test. It provides the assumptions, data arrangement, hypotheses, test statistic, decision rules, and example application of both tests for categorical paired and unpaired data. Fisher's exact test is used for 2x2 contingency tables when sample sizes are small, while McNemar's test analyzes paired nominal data to evaluate hypotheses about proportions between pairs.
Chi‑square Test and its Application in Hypothesis Testing
Rakesh Rana, Richa Singhal
Statistical Section, Central Council for Research in Ayurvedic Sciences, Ministry of AYUSH, GOI, New Delhi, India
Chi‑square Test and its Application in Hypothesis Testing
Rakesh Rana, Richa Singhal
Statistical Section, Central Council for Research in Ayurvedic Sciences, Ministry of AYUSH, GOI, New Delhi, India
1. The standard deviation of the diameter at breast height, or DBH.docxpaynetawnya
1. The standard deviation of the diameter at breast height, or DBH, of the slash pine tree is less than one inch. Identify the Type I error. (Points : 1)
Fail to support the claim σ < 1 when σ < 1 is true.
Support the claim μ < 1 when μ = 1 is true.
Support the claim σ < 1 when σ = 1 is true. Fail to support the claim μ < 1 when μ < 1 is true.
1a. The EPA claims that fluoride in children's drinking water should be at a mean level of less than 1.2 ppm, or parts per million, to reduce the number of dental cavities. Identify the Type I error. (Points : 1)
Fail to support the claim σ < 1.2 when σ < 1.2 is true.
Support the claim μ < 1.2 when μ = 1.2 is true.
Support the claim σ < 1.2 when σ = 1.2 is true.
Fail to support the claim μ < 1.2 when μ < 1.2 is true.
2. Biologists are investigating if their efforts to prevent erosion on the bank of a stream have been statistically significant. For this stream, a narrow channel width is a good indicator that erosion is not occurring. Test the claim that the mean width of ten locations within the stream is greater than 3.7 meters. Assume that a simple random sample has been taken, the population standard deviation is not known, and the population is normally distributed. Use the following sample data:
3.3 3.3 3.5 4.9 3.5 4.1 4.1 5 7.3 6.2
What is the P-value associated with your test statistic? Report your answer with three decimals, e.g., .987 (Points : 1)
2a. Medical researchers studying two therapies for treating patients infected with Hepatitis C found the following data. Assume a .05 significance level for testing the claim that the proportions are not equal. Also, assume the two simple random samples are independent and that the conditions np ≥ 5 and nq ≥ 5 are satisfied.
Therapy 1
Therapy 2
Number of patients
39
47
Eliminated Hepatitis
20
13
C infection
Construct a 95% confidence interval estimate of the odds ratio of the odds for having Hepatitis C after Therapy 1 to the odds for having Hepatitis C after Therapy 2. Give your answer with two decimals, e.g., (12.34,56.78) (Points : 0.5)
3. Researchers studying sleep loss followed the length of sleep, in hours, of 10 individuals with insomnia before and after cognitive behavioral therapy (CBT). Assume a .05 significance level to test the claim that there is a difference between the length of sleep of individuals before and after CBT. Also, assume the data consist of matched pairs, the samples are simple random samples, and the pairs of values are from a population having a distribution that is approximately normal.
Individual
1
2
3
4
5
6
7
8
9
10
Before
6
5
4
5
3
4
5
3
4
2
CBT
After
8
8
7
6
7
6
6
5
7
5
CBT
Construct a 95% confidence interval estimate of the mean difference between the lengths of sleep. Give your answer with two decimals, e.g., (12.34,56.78) (Points : 0.5)
3a. Scientists, researching large woody debris (LWD), surveyed the number of LWD ...
Please Subscribe to this Channel for more solutions and lectures
http://www.youtube.com/onlineteaching
Chapter 9: Inferences from Two Samples
9.4: Two Variances or Standard Deviations
36086 Topic Discussion3Number of Pages 2 (Double Spaced).docxrhetttrevannion
36086 Topic: Discussion3
Number of Pages: 2 (Double Spaced)
Number of sources: 1
Writing Style: APA
Type of document: Essay
Academic Level:Master
Category: Psychology
Language Style: English (U.S.)
Order Instructions: Attached
I will upload the instructions
Reference/Module
Learning Objectives
•Explain what the x2 goodness-of-fit test is and what it does.
•Calculate a x2 goodness-of-fit test.
•List the assumptions of the x2 goodness-of-fit test.
•Calculate the x2 test of independence.
•Interpret the x2 test of independence.
•Explain the assumptions of the x2 test of independence.
The Chi-Square (x2) Goodness-of-Fit test: What It Is and What It Does
The chi-square (x2) goodness-of-fit test is used for comparing categorical information against what we would expect based on previous knowledge. As such, it tests what are called observed frequencies (the frequency with which participants fall into a category) against expected frequencies (the frequency expected in a category if the sample data represent the population). It is a nondirectional test, meaning that the alternative hypothesis is neither one-tailed nor two-tailed. The alternative hypothesis for a x2 goodness-of-fit test is that the observed data do not fit the expected frequencies for the population, and the null hypothesis is that they do fit the expected frequencies for the population. There is no conventional way to write these hypotheses in symbols, as we have done with the previous statistical tests. To illustrate the x2 goodness-of-fit test, let's look at a situation in which its use would be appropriate.
chi-square (x2) goodness-of-fit test A nonparametric inferential procedure that determines how well an observed frequency distribution fits an expected distribution.
observed frequencies The frequency with which participants fall into a category.
expected frequencies The frequency expected in a category if the sample data represent the population.
Calculations for the x2 Goodness-of-Fit Test
Suppose that a researcher is interested in determining whether the teenage pregnancy rate at a particular high school is different from the rate statewide. Assume that the rate statewide is 17%. A random sample of 80 female students is selected from the target high school. Seven of the students are either pregnant now or have been pregnant previously. The χ2goodness-of-fit test measures the observed frequencies against the expected frequencies. The observed and expected frequencies are presented in Table 21.1.
TABLE 21.1Observed and expected frequencies for χ2 goodness-of-fit example
FREQUENCIES
PREGNANT
NOT PREGNANT
Observed
7
73
Expected
14
66
As can be seen in the table, the observed frequencies represent the number of high school females in the sample of 80 who were pregnant versus not pregnant. The expected frequencies represent what we would expect based on chance, given what is known about the population. In this case, we would expect 17% of the females to be pregnant .
Explore our infographic on 'Essential Metrics for Palliative Care Management' which highlights key performance indicators crucial for enhancing the quality and efficiency of palliative care services.
This visual guide breaks down important metrics across four categories: Patient-Centered Metrics, Care Efficiency Metrics, Quality of Life Metrics, and Staff Metrics. Each section is designed to help healthcare professionals monitor and improve care delivery for patients facing serious illnesses. Understand how to implement these metrics in your palliative care practices for better outcomes and higher satisfaction levels.
CHAPTER 1 SEMESTER V PREVENTIVE-PEDIATRICS.pdfSachin Sharma
This content provides an overview of preventive pediatrics. It defines preventive pediatrics as preventing disease and promoting children's physical, mental, and social well-being to achieve positive health. It discusses antenatal, postnatal, and social preventive pediatrics. It also covers various child health programs like immunization, breastfeeding, ICDS, and the roles of organizations like WHO, UNICEF, and nurses in preventive pediatrics.
The Importance of Community Nursing Care.pdfAD Healthcare
NDIS and Community 24/7 Nursing Care is a specific type of support that may be provided under the NDIS for individuals with complex medical needs who require ongoing nursing care in a community setting, such as their home or a supported accommodation facility.
India Clinical Trials Market: Industry Size and Growth Trends [2030] Analyzed...Kumar Satyam
According to TechSci Research report, "India Clinical Trials Market- By Region, Competition, Forecast & Opportunities, 2030F," the India Clinical Trials Market was valued at USD 2.05 billion in 2024 and is projected to grow at a compound annual growth rate (CAGR) of 8.64% through 2030. The market is driven by a variety of factors, making India an attractive destination for pharmaceutical companies and researchers. India's vast and diverse patient population, cost-effective operational environment, and a large pool of skilled medical professionals contribute significantly to the market's growth. Additionally, increasing government support in streamlining regulations and the growing prevalence of lifestyle diseases further propel the clinical trials market.
Growing Prevalence of Lifestyle Diseases
The rising incidence of lifestyle diseases such as diabetes, cardiovascular diseases, and cancer is a major trend driving the clinical trials market in India. These conditions necessitate the development and testing of new treatment methods, creating a robust demand for clinical trials. The increasing burden of these diseases highlights the need for innovative therapies and underscores the importance of India as a key player in global clinical research.
Health Education on prevention of hypertensionRadhika kulvi
Hypertension is a chronic condition of concern due to its role in the causation of coronary heart diseases. Hypertension is a worldwide epidemic and important risk factor for coronary artery disease, stroke and renal diseases. Blood pressure is the force exerted by the blood against the walls of the blood vessels and is sufficient to maintain tissue perfusion during activity and rest. Hypertension is sustained elevation of BP. In adults, HTN exists when systolic blood pressure is equal to or greater than 140mmHg or diastolic BP is equal to or greater than 90mmHg. The
How many patients does case series should have In comparison to case reports.pdfpubrica101
Pubrica’s team of researchers and writers create scientific and medical research articles, which may be important resources for authors and practitioners. Pubrica medical writers assist you in creating and revising the introduction by alerting the reader to gaps in the chosen study subject. Our professionals understand the order in which the hypothesis topic is followed by the broad subject, the issue, and the backdrop.
https://pubrica.com/academy/case-study-or-series/how-many-patients-does-case-series-should-have-in-comparison-to-case-reports/
ICH Guidelines for Pharmacovigilance.pdfNEHA GUPTA
The "ICH Guidelines for Pharmacovigilance" PDF provides a comprehensive overview of the International Council for Harmonisation of Technical Requirements for Pharmaceuticals for Human Use (ICH) guidelines related to pharmacovigilance. These guidelines aim to ensure that drugs are safe and effective for patients by monitoring and assessing adverse effects, ensuring proper reporting systems, and improving risk management practices. The document is essential for professionals in the pharmaceutical industry, regulatory authorities, and healthcare providers, offering detailed procedures and standards for pharmacovigilance activities to enhance drug safety and protect public health.
Defecation
Normal defecation begins with movement in the left colon, moving stool toward the anus. When stool reaches the rectum, the distention causes relaxation of the internal sphincter and an awareness of the need to defecate. At the time of defecation, the external sphincter relaxes, and abdominal muscles contract, increasing intrarectal pressure and forcing the stool out
The Valsalva maneuver exerts pressure to expel faeces through a voluntary contraction of the abdominal muscles while maintaining forced expiration against a closed airway. Patients with cardiovascular disease, glaucoma, increased intracranial pressure, or a new surgical wound are at greater risk for cardiac dysrhythmias and elevated blood pressure with the Valsalva maneuver and need to avoid straining to pass the stool.
Normal defecation is painless, resulting in passage of soft, formed stool
CONSTIPATION
Constipation is a symptom, not a disease. Improper diet, reduced fluid intake, lack of exercise, and certain medications can cause constipation. For example, patients receiving opiates for pain after surgery often require a stool softener or laxative to prevent constipation. The signs of constipation include infrequent bowel movements (less than every 3 days), difficulty passing stools, excessive straining, inability to defecate at will, and hard feaces
IMPACTION
Fecal impaction results from unrelieved constipation. It is a collection of hardened feces wedged in the rectum that a person cannot expel. In cases of severe impaction the mass extends up into the sigmoid colon.
DIARRHEA
Diarrhea is an increase in the number of stools and the passage of liquid, unformed feces. It is associated with disorders affecting digestion, absorption, and secretion in the GI tract. Intestinal contents pass through the small and large intestine too quickly to allow for the usual absorption of fluid and nutrients. Irritation within the colon results in increased mucus secretion. As a result, feces become watery, and the patient is unable to control the urge to defecate. Normally an anal bag is safe and effective in long-term treatment of patients with fecal incontinence at home, in hospice, or in the hospital. Fecal incontinence is expensive and a potentially dangerous condition in terms of contamination and risk of skin ulceration
HEMORRHOIDS
Hemorrhoids are dilated, engorged veins in the lining of the rectum. They are either external or internal.
FLATULENCE
As gas accumulates in the lumen of the intestines, the bowel wall stretches and distends (flatulence). It is a common cause of abdominal fullness, pain, and cramping. Normally intestinal gas escapes through the mouth (belching) or the anus (passing of flatus)
FECAL INCONTINENCE
Fecal incontinence is the inability to control passage of feces and gas from the anus. Incontinence harms a patient’s body image
PREPARATION AND GIVING OF LAXATIVESACCORDING TO POTTER AND PERRY,
An enema is the instillation of a solution into the rectum and sig
Leading the Way in Nephrology: Dr. David Greene's Work with Stem Cells for Ki...Dr. David Greene Arizona
As we watch Dr. Greene's continued efforts and research in Arizona, it's clear that stem cell therapy holds a promising key to unlocking new doors in the treatment of kidney disease. With each study and trial, we step closer to a world where kidney disease is no longer a life sentence but a treatable condition, thanks to pioneers like Dr. David Greene.
1. Chi Square distribution and
analysis of frequency
Dr Inn Kynn Khaing
M.B.,B.S, MPH, MSc (Healthcare Administration) (Japan)
Lecturer, Department of Biostatistics
3. The chi-square test is not an appropriate method of analysis if minimum
expected frequency requirements are not met.
For example : if n is less than 20 or if n is between 20 and 40 and one of the
expected frequencies is less than 5, the chi-square test should be avoided.
It is called exact because, if desired, it permits us to calculate the exact
probability of obtaining the observed results or results that are more
extreme.
Assumptions
1. The data consist of A sample observations from population 1 and B
sample observations from population 2.
2. The samples are random and independent.
3. Each observation can be categorized as one of two mutually exclusive
types.
3
4. Data Arrangement
When we use the Fisher exact test, we arrange the data in the form of a
2 x 2 contingency table.
We arrange the frequencies in such a way that A > B and choose the
characteristic of interest so that a/A > b/B. Some theorists believe that
Fisher’s exact test is appropriate only when both marginal totals of Table
are fixed by the experiment.
4
5. 5
Hypotheses
Two-sided
H0: The proportion with the characteristic of interest is the same in
both populations; i.e, p1 = p2
HA: The proportion with the characteristic of interest is not the same
in both populations; p1 ≠ p2
One-sided
H0: The proportion with the characteristic of interest in population 1
is less than or the same as the proportion in population 2; p1 ≤ p2.
HA: The proportion with the characteristic of interest is greater in
population 1 than in population 2; p1 > p2.
6. 6
Test Statistic
The test statistic is b, the number in sample 2 with the characteristic of
interest.
Decision Rule
Two-sided test
If the observed value of b is ≤ the integer in a given column, reject H0 at a
level of significance equal to twice the significance level shown at the top
of that column.
For example, suppose A = 8, B = 7, a = 7, and the observed value of b is 1.
We can reject the null hypothesis at the 2(.05)=.10, the 2(0.025)= 0.05, and
the 2 (0.01)=0.02 levels of significance, but not at the 2 (0.005)=.01 level
7. 7
Decision Rule
2. One-sided test
If the observed value of b is less than or equal to the integer in a given
column, reject H0 at the level of significance shown at the top of that
column.
For example, suppose A = 16, B = 8, a = 4, and the observed value of b is 3.
We can reject the null hypothesis at the .05 and .025 levels of significance,
but not at the .01 or .005 levels.
8. The purpose of a study was to evaluate the long-term efficacy of taking
indinavir/ritonavir twice a day in combination with two nucleoside reverse
transcriptase inhibitors among HIV-positive subjects who were divided into two
groups. Group 1 consisted of patients who had no history of taking protease
inhibitors. Group 2 consisted of patients who had a previous history taking a
protease inhibitor. Table shows whether these subjects remained on the
regimen for the 120 weeks of follow-up.
We wish to know if we may conclude that patients classified as group 1 have a
lower probability than subjects in group 2 of remaining on the regimen for 120
weeks.
8
9. Solution:
1. Data. The data are rearranged to conform to the layout of given table.
Remaining on the regimen is the characteristic of interest.
2. Assumptions. We presume that the assumptions for application of the Fisher
exact test are met.
9
We arrange the frequencies in such a way that A > B and
choose the characteristic of interest so that a/A > b/B.
10. Solution:
3. Hypotheses.
H0: The proportion of subjects remaining 120 weeks on the regimen in a
population of patients classified as group 2 is the same as or less than the
proportion of subjects remaining on the regimen 120 weeks in a population
classified as group 1.
HA: Group 2 patients have a higher rate than group 1 patients of remaining
on the regimen for 120 weeks.
4. Test statistic. The test statistic is the observed value of b as shown in table.
5. Distribution of test statistic. We determine the significance of b by
consulting Appendix Table J.
6. Decision rule. Suppose we let α=.05. The decision rule, then, is to reject H0 if
the observed value of b is equal to or less than 1, the value of b in Table J for
A = 12, B = 9, a = 8, and α=.05.
10
11. Solution:
7. Calculation of test statistic. The observed value of b, as shown in table is 2.
8. Statistical decision. Since 2 > 1, we fail to reject H0.
9. Conclusion. Since we fail to reject H0, we conclude that the null hypothesis
may be true. That is, it may be true that the rate of remaining on the
regimen for 120 weeks is the same or less for the PI experienced group
compared to the PI naive group.
10. p value. We see in Table J that when A = 12, B = 9, a = 8, the value of b = 2
has an exact probability of occurring by chance alone, when H0 is true,
greater than .05. p>0.05
11
12. Test for paired nominal data
When categorical data are paired, the McNemar test is the appropriate test
When data are paired and the outcome of interest is a proportion, the
McNemar Test is used to evaluate hypotheses about the data.
Developed by Quinn McNemar in 1947
Sometimes called the McNemar Chi-square test because the test statistic
has a Chi-square distribution
1st subject of pair
Variable 1
Variable 2 0 1 Total
2nd subject of pair
0 e f
e + f
1 g h
g + h
Total e + g f + h n 12
13. Example:
Over a period of a 6 months the registrar selected every patient with this
disorder and paired them off as far as possible by reference to age, sex,
and frequency of ulceration. Finally she had 108 patients in 54 pairs. To
one member of each pair, chosen by the toss of coin, she gave treatment
A, which she and her colleagues in the unit had hitherto regarded as the
best; to the other member she gave the new treatment, B. Both forms of
treatments are local applications, and they cannot be made to look alike.
Consequently to avoid bias in the assessment of the results a colleague
recorded the results of treatment without knowing which patient in each
pair had which treatment.
13
14. Member of pair
receiving treatment A
Member of pair
receiving treatment B
Pairs of patients
Responded Responded 16
Responded Did not respond 23
Did not respond Responded 10
Did not respond Did not respond 5
Total 54
1st subject of pair
Responded Did not respond
2nd subject of
pair
Responded 16 10
Did not respond 23 5
Total e + g f + h 14
15. 1st subject of pair
Responded Did not respond
2nd subject of
pair
Responded 16 (e) 10 (f)
Did not respond 23 (g) 5 (h)
X2 = (f-g)2/ (f + g) with 1 d.f
X2 = (10-23)2/ (10 + 23) = 5.12
Or with a continuity correction
X2 = ( ǀf-gǀ -1 )2/ (f + g) with 1 d.f
X2 = ( ǀ10-23ǀ -1 )2/ (10 + 23) = 4.36
Both X2 values 0.02 < P < 0.05
Conclusion:
Treatment A gave significantly
better results than treatment B.
15
16. The sampling distribution of the McNemar statistic is a Chi-square distribution.
For a test with alpha = 0.05, the critical value for the McNemar statistic = 3.84.
The null hypothesis is not rejected if the McNemar statistic < 3.84.
The null hypothesis is rejected if the McNemar statistic > 3.84.
16
17. For ( r x 1) or (1 x c) table : df = k-1 where k = number of categories
For many assumptions : df = k=r where r = number of restrictions
For (r x c) table : df = (r-1) (c-1) where r = rows, c = column
The most simple and commonly applied table is 2 x 2 table which has two rows
and columns.
17
19. 4. In a study of relationship between gender and smoking habit, 100 couples had been
interviewed. After controlling age and socioeconomic status, it was found that
Both male and female smokers 24 couples
Neither male nor female smokers 40 couples
Male but Not female smokers 26 couples
Not male but female smokers 10 couples
Construct a cross-table from the above data.
Analyze these data by using McNemar’s test.
Find 95% confidence limits of the difference in proportions.
19