Lecture 9 t-test for one sample.pptx

shakirRahman10 85 views 34 slides Jul 16, 2023
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About This Presentation

Objectives:
Generate of t-test.
Learn about the assumptions of t-test.
Calculate t-test.
Construct the confidence interval for the population mean.
Recall steps for z test:
1. State null and alternative hypothesis.
2. Determine the level of significance
3. Apply test statistics.
4. Identify critica...


Slide Content

2 Shakir Rahman BScN , MScN , MSc Applied Psychology, PhD Nursing (Candidate) University of Minnesota USA. Principal & Assistant Professor Ayub International College of Nursing & AHS Peshawar Visiting Faculty Swabi College of Nursing & Health Sciences Swabi Nowshera College of Nursing & Health Sciences Nowshera

By the end of the session, students will be able to: Generate of t-test . Learn about the assumptions of t-test. Calculate t-test. Construct the confidence interval for the population mean. 3

Descriptive Infe r e n tia l St a t i s t i cs Es timat i on H y p o t h e s i s testing Point Es timate Interval Es timate 4

1. State null and alternative hypothesis. 5 2. Determine the level of significance Apply test statistics. Identify critical region/ p-value. Interpret the result Z-Test

Clue: Look at the z test formula. 6

 When population standard deviation is known or sample is large enough that sample population deviation will represent population’s standard deviation. We can used z-test or standard normal distribution.  What do we do if population standard deviation is not known and the sample size is less than 30? 7

Also known as student’s test. Identified by William Goset. 8

Bell Shaped It is symmetric around the mean. Mean, median, and the mode are plot at zero which is at the center of bell shape curve. The curve does not touch the x-axis. 9

T- distribution is associated with degrees of freedom. Degree of freedom is (n-1) and is associated with sample size. Degree of freedom are the number of values that are free to vary after a sample statistics has been computed . It tells the research which t curve to use. With the increase in sample size, the t distribution approaches the standard normal distribution. 11

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The average score of 5 individuals is 16 points. The 4 out of 5 values are free to vary. But once the 4 values are determined, the fifth value must be a specific number to yield 16 points as mean of all 5 values. 25 points. 15 points. 10 points. 25points. 5 points. 13

Standard deviation of population is unknown. Use the s (standard deviation). If sample size less than 30 so normal distribution should be ensured. 14

State null and alternative hypothesis. Determine the level of significance Apply test statistics. Identify critical region/ p-value. Interpret the result 15

Calculate critical value of t with alpha 0.01 and d.f.= 21for left tail. Calculate value for alpha as 0.1 with d.f. 17 for a two tailed t-test. Find the critical value for alpha 0.05 with d.f.= 28 for right tailed t-test. 16

 -2.518  +1.740, -1.740  +1.701 18

A nurse researcher claims that the mean number of infections in a week at a hospital in a country is 350 cases. A random sample of 12 weeks had a mean number of 358 cases. The standard deviation of sample is 16. Test the claim at alpha 0.05 that the average is higher than 350. 19

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Step 4: Reject Ho if t cal > t tabα, d.f. t cal > t tab 0.05,11. 1.732 < 1.796 (Fail to reject ) 21

1.796 23 1.796

Step V: 02 - O c t - 20 24 Since t cal > t tabα, d.f. and does not fall in the rejection region, so we fail to reject Ho at 5 % level of significance and there is not enough evidence to conclude that mean infection rate per week is greater than 350.

The human resource manager claims that the average salary of teachers in nursing school in a country is found to be different from 60$ a day. A random sample from 8 nursing schools were selected, and the daily salaries (in $) are mentioned below. The value of sample standard deviation is 5.08. Is this enough evidence to support the manager’s claim at alpha 0.1? 60 56 60 55 70 55 60 55 25 Bluman (2012)

Step 1: Ho: μ = 60$ Ha: μ ≠ 60$ Step 2: At alpha: 0.10 and d.f. 7, critical value is 1.895 t tabα, d.f.=t tab 0.1,7= 1.895 26

Step 3: To compute test value, the mean and standard deviation must be found 28

58.88-60/5.08 /√ 8 -0.62 Step 4: Do not reject null hypothesis since -0.624 does not fall in the critical region. R e j ect Ho if t cal> t tab or t cal< -t tab - 1.8 9 5 - 0.6 2 4 +1.8 9 5 29

Interpretation: There is not enough evidence to support manager’s claim that the average salary of nursing faculty in a country is different from 60$ a day at 10% level of significance . 30

Bluman (2012). Elementary Statistics: A Step by Step Approach (8 th .). McGraw Hill. Daniel (2014). Biostatistics: Basic Concepts and Methodology for the Health Sciences. New York: John Wiley & Sons. 32

Acknowledgements Dr Tazeen Saeed Ali RM, RM, BScN, MSc ( Epidemiology & Biostatistics), Phd (Medical Sciences), Post Doctorate (Health Policy & Planning) Associate Dean School of Nursing & Midwifery The Aga Khan University Karachi. Kiran Ramzan Ali Lalani BScN, MSc Epidemiology & Biostatistics (Candidate) Registered Nurse (NICU) Aga Khan University Hospital