Probability theoryThe probability of an event may be determined empirically (by observation) or mathematically (using probability theory). Probability theory is fundamentally important to inferential statistical analysis.

Predicting population parameters from sample data is based on the assumption that the sample data are ‘typical’ of the population data.

The laws of probability govern just how typical the data are 1 Oct 2007 - Probability theory is fundamentally important to inferential statistical analysis. This is often the case, even when the sample means are significantly different. The study comparing two hypotensive agents could be designed differently. Royal College of Anaesthetists College of Anaesthetists of Ireland .

For example, we may toss a coin 20 times to determine the likelihood of obtaining heads on a single throw. Common sense tell us that, provided the coin is unbiased with heads just as likely to fall as tails, the ratio of heads:tails should be 1:1 and therefore the ‘expected’ outcome after 20 tosses would be 10 heads.

However, the actual outcome may well be different. If we were to repeat the experiment by tossing the coin 1000 times, it is likely that the ratio heads:tails would be very close to 1:1 and if the coin was tossed an infinite number of times, the ratio would be exactly 1:1.

We may consider the population of interest in this scenario to be the outcome of an infinite number of coin tosses. A sample drawn from this population is an experiment in which the coin is tossed a finite number of times.

Returning to the experiment in which the coin is tossed 20 times, probability theory may be used to determine mathematically the likelihood of obtaining any combination of heads and tails. The probability (P) of an event occurring is an expression of the relative frequency that the event occurs in an infinite number of trials.

P ranges from 0 (the event never occurs) to 1 (the event always occurs). Let us consider eye colour and suppose that eyes are either blue, brown, grey, or green.

The probability that an individual's eyes are blue is given by the expression P(blue). The four categories are exhaustive and mutually exclusive. The probability that an individual's eyes are coloured either blue or brown is given by the expression P(blue or brown) = Additionally, if the events A and B are exhaustive, P(A or B) = 1.

If two events are not mutually exclusive, they be independent (the probability of one event occurring is not affected by whether or not another event occurs) Advanced Probability Theory Stats Homework, assignment and Project Help, Advanced Probability Theory Assignment Math Tools: Online Math tools to help with homework or studying. StudyDaddy is the place where you can get easy online Statistics homework. Spend less money when buying from our drugstore..

If two events A and B are not mutually exclusive but are independent, then: The binomial distributionReturning to the coin tossing experiment, if we toss the coin three times, the following outcomes are possible: TTT, TTH, THT, HTT, THH, HTH, HHT, HHH. If we assume that the probability of obtaining either a head or a tail is equally likely on each toss, then each of the eight possible combinations listed above is equally likely to occur with a probability of 1/8.

Accordingly, the probability of obtaining no heads in three trials is 1/8, one head is 3/8, two heads is 3/8, and three heads is 1/8. The same exercise could be repeated by tossing the coin a 100 times (Fig.

1Binomial distribution: the probability of obtaining heads after tossing a coin 100 times. The area under the curve (which follows a normal distribution) = 1.

1Binomial distribution: the probability of obtaining heads after tossing a coin 100 times.

The area under the curve (which follows a normal distribution) = 1. It is observed that as n increases, the number of heads obtained tends towards a normal distribution.

This type of distribution of two independent outcomes is termed a binomial distribution In fact, we will study probability theory based on the theory number of elements in the sample space S. For a discrete case, the probability of an event A can be .

In this particular example, the probability of one of the outcomes (heads) is 0.

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obtaining sixes after throwing a die a 100 times.

Generally, a binomial distribution may be used to describe any situation where there are n independent trials with two mutually exclusive, independent outcomes, the outcome of interest occurring with a probability of p on each trial. It follows a normal distribution provided n is reasonably large and p does not take too extreme a value (close to 0 or 1).

It can be shown mathematically that: The mean of a binomial distribution = np The standard deviation of a binomial distribution = Statistical inferenceOne group of interval dataSometimes, we may wish to analyse just one dataset. For example, we may wish to infer a population parameter (e.

Or, we may wish to determine whether the mean (or median) of a sample dataset differs from either a known value (e. Estimation of the population mean from sample dataSuppose that we want to know the average IQ of all UK trainees in anaesthesia. We will assume that IQ follows a normal distribution.

As testing the entire population is impractical, we decide to test a random sample of 200 trainees.

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How accurate is the sample estimate of the population mean (the mean IQ of all UK trainees in anaesthesia)?If we were to repeat the same investigation numerous times, we would obtain a series of sample means that would follow a normal distribution Case Studies Vipul Vaibhaw, B.Sc Mathematics & Statistics, Fergusson College, Pune (2017) How large must n be to make this probability greater than 50%? 4. What are the most interesting open problems in probability theory?.

This is the central limit theorem and it applies even when the population data are not normally distributed.

The mean of this sampling distribution is equal to the population mean . The standard deviation of the sampling distribution equals sd/√n i.

As we do not know the population standard deviation, the sample standard deviation is used instead. From previous discussions, we know that ∼95% of a sample of normally distributed data lies within ± 1.

Thus, the 95% confidence interval for the population mean IQ is given by the expression x̄ − 1.

96 × (SD/√t-testSuppose now that we wish to know how the average IQ of UK trainees in anaesthesia compares with the ‘known’ data for the UK adult population as a whole. The estimate of the UK population data for trainees in anaesthesia obtained in the above investigation is used and compared with the known (published) data on the IQ of the adult UK population as a whole using a one-sample t-test. t-value analogous to the z-value previously discussed in relation to the standard normal distribution is calculated according to the equation: t = (x̄ − )/sem.

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In fact, the t-distribution comprises a family of curves depending on sample size; the t-distribution used for a given sample size is specified by the number of degrees of freedom (equal to n − 1). Wilcoxon rank sum testIn this test, a sample median is compared against a known or hypothetical population median in a non-parametric distribution.

Each sample datum is assigned a rank depending on how far it is from the median. Datum values lower than the median are given negative values.

All of these signed ranks are summed to produce a W-value. If the null hypothesis is true, W is near to zero.

Comparing two groups of interval dataIn clinical studies, we often want to compare two sample groups. Two key criteria must be specified: are the data normally distributed and are the data paired?Unpaired (independent) normally distributed data: Student's unpaired two-sample t-testFor example, the efficacy of a new hypotensive drug A may be compared with an established drug B.

The study has nx̄A and nB do not have to be equal). We need to calculate the difference between the two sample means and the standard error of this difference between the two means, from which we can calculate a confidence interval for the difference between them.

For Student's t-test to be valid, the standard deviations of both groups must be similar. This is often the case, even when the sample means are significantly different. Most statistics software programs will routinely check that this is true.

If the two sample standard deviations are observed to be unequal, Welch's correction to Student's t-test should be applied.

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Instead of having two independent groups, all of the patients recruited could be treated with one of the two study drugs (decided upon by random allocation) and the effect of treatment measured after a period of stabilization.

Drug treatment is then stopped and after a washout period during which arterial blood pressure levels return to baseline levels, treatment with the other drug is commenced and its effect determined In fact, we will study probability theory based on the theory number of elements in the sample space S. For a discrete case, the probability of an event A can be .

This type of study in which all of the subjects receive both drugs under investigation is called a crossover study. The design of a crossover study involves the analysis of matched pairs of data. In this situation, the appropriate statistical test is Student's paired t-test.

Instead of analysing the data of two pooled groups, the effects of drug treatment on each individual in either arm of the study is separately analysed. As we shall see later, this form of analysis is more powerful.

Non-parametric interval dataStudent's t-test is not used for data that does not follow a normal distribution. The analogous statistical test to the unpaired t-test is the Mann–Witney U-test; the analogous test to the paired t-test is the Wilcoxon matched pairs test.

Both tests analyse the data by comparing the medians rather than the means, and by considering the data as rank order values rather than absolute values. Three or more groups of interval dataThe t-tests and their non-parametric equivalents are only used to compare two groups. When there are three or more groups under investigation, the appropriate test for normally distributed interval data is analysis of variance (anova).

If anova testing suggests the groups are different, we are usually interested in knowing between which specific groups the differences exist.

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This approach is inherently more robust than simply performing three two-sample t-tests as we only proceed to compare pairs of data once we have evidence of a significant difference between all of the study groups.

anova may also be used to compare just two study groups, when it is equivalent to Student's unpaired t-test 3 Jul 2015 - But when you get probability questions in your GRE and GMAT exam syllabus, Understanding the basic rules and formulas of probability will help you score high is probable; the likelihood of something happening or being the case'. Example: In a class, 40% of the students study math and science..

When three or more normally distributed datasets are matched, the repeated measures anova test is equivalent to Student's paired t-test. For data that is not normally distributed, the Kruskal–Wallis anova by ranks test is used for independent groups and the Friedman test for matched datasets.

Categorical dataWhen data are classified into groups, either the Fisher exact or the 2 test is used to determine whether the sample proportions in each group are significantly different. A contingency table containing the data is produced as previously described.

A 2 × 2 contingency table is best analysed using the Fisher exact test. Although the 2 test can also be used for analysis of 2 × 2 tables (when Yates' correction is usually applied), it gives a less accurate result. It was used in the past, as it is easier to calculate than the Fisher exact test.

However, with the widespread availability of computer software packages for statistical analysis, the Fisher exact test is preferable. Two statistical measures of the relative likelihood of an event or outcome occurring in two sample groups may be defined: the relative risk and odds ratio. When the data in a contingency table relate to a prospective study, both these measures may be calculated.

Only the odds ratio may be calculated for retrospective case–control studies College of Engineering The basic concepts will be illustrated by use of case studies. 310, 311. Elements of Probability and Mathematical Statistics I and II. treatment of experimental data: normal sampling theory; confidence intervals and .

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Table 1The relative risk of obtaining a given outcome after one intervention compared with another is equal to the ratio of the observed risk of the outcome after the first intervention divided by the observed risk of the outcome after the second intervention. The key factor in calculating relative risk is knowing the actual number of individuals at risk in each group.

In a prospective study, the number of patients at risk of an outcome is known, whereas in a retrospective study, the outcome is the starting point and the number of patients at risk is not known. The odds ratio is defined as the odds of the outcome of interest occurring after the first intervention divided by the odds of the outcome of interest occurring after the second intervention, where there are two mutually exclusive outcomes.

The odds of an event or outcome in each of the two study groups = p/(1 −