**ST2004 Statistical Inference Assignment Sample NUIG Ireland**

ST2004 Statistical Inference is an introductory course, aimed at first-year students of statistics and related disciplines. The course covers the basic principles of estimation and hypothesis testing for both point and interval estimations.

The most important thing you need to know about ST2004 is that it provides a solid foundation for understanding more advanced courses in statistical inference. If you’re planning to take additional courses in statistics, then ST2004 is a course you won’t want to miss.

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In this section, we are describing some assigned tasks. These are:

**Assignment Task 1: Derive a likelihood function for random samples from a probability model under more complex sampling schemes, eg mixed populations, and censoring.**

Chances are you encounter random sampling schemes frequently and may not realize it. For example, whenever you take a survey, the sample is usually drawn randomly from the population to arrive at more reliable conclusions that can be generalized to the larger group. Even something as simple as flipping a coin can be thought of as a form of random sampling, where each flip corresponds to a draw from a population of size two (heads or tails).

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Under more complex sampling schemes, such as mixed populations or censoring, deriving the likelihood function can get tricky. However, the basic idea is still the same: we want to figure out how likely it is that our data came from a particular probability model. This information can then be used to make inferences about the underlying population.

For example, suppose we want to study the effects of a new drug on blood pressure. We could randomly select a sample of people from the population and measure their blood pressure before and after taking the drug. However, this would not be an ideal way to conduct the study, since there would be no way to control for confounding factors, such as age, weight, or preexisting conditions.

A better way to conduct the study would be to randomly assign people to two groups: one group that takes the drug and one group that doesn’t. This is known as a randomized controlled trial (RCT). By randomly assigning people to groups, we can be more confident that any differences in blood pressure between the two groups are due to the drug and not to other factors.

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**Assignment Task 2: Calculate simple unbiased estimators and calculate optimal combinations of estimators.**

There are many ways to calculate simple unbiased estimators. One method is to use the method of moments. This technique relies on the first and second moment of an unbiased estimator being equal to the first and second moment of the data.

Another common method is maximum likelihood estimation. This approach finds estimates by searching for values that maximize the likelihood function. These values are usually found by taking derivatives of the log-likelihood function and setting them equal to zero.

Optimal combinations of estimators can be calculated using a variety of methods, including linear programming or dynamic programming. In general, these methods aim to find a combination of estimators that minimizes some cost functions while still providing accurate results. Linear programming is a more general method that can be applied to a wide variety of problems, while dynamic programming is specifically designed for problems with an underlying Markov process.

**Assignment Task 3: Find maximum likelihood estimators by solving the score equation and obtain an estimate of precision based on observed and expected information.**

Maximum likelihood estimators can be found by solving the score equation, and the precision of the estimate can be based on observed and expected information.

Observed information is the information in the data, while expected information is the information that would be expected if the model were true. This expected information can be computed by taking the derivative of the log-likelihood concerning each parameter and setting it to zero. The maximum likelihood estimate is then the value of the parameter that results in a maximum of the log-likelihood function.

The precision of an estimate can also be determined by its standard error, which is a measure of how much variability there is in estimates from repeated samples. The standard error can be computed as the square root of the variance of the estimate. The variance of the estimate is the sum of the squares of the bias and the standard error.

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**Assignment Task 4: Find confidence intervals for simple problems using pivotal quantities.**

We can derive confidence intervals for population parameters using pivotal quantities. A pivotal quantity is a function of the sample data that has a known sampling distribution. In other words, it’s a statistic that we can use to make inferences about the population parameter.

Let’s say we want to find a 95% confidence interval for the population mean μ. We can do this by finding the 2.5th and 97.5th percentiles of the sampling distribution of the sample mean x̄. These values will form our lower and upper bounds, respectively.

Recall that the formula for the standard error of the mean is σ/√n, where σ is the population standard deviation and n is the sample size. This means that the variance of the sampling distribution of the sample mean is σ^2/n.

We can use this formula to find the 2.5th and 97.5th percentiles of the sampling distribution of the sample mean. The 2.5th percentile is simply μ-1.96σ/√n, and the 97.5th percentile is μ+1.96σ/√n. This gives us a 95% confidence interval for the population mean μ of μ-1.96σ/√n to μ+1.96σ/√n.

**Assignment Task 5: Calculate the size and power function for a given test procedure.**

To calculate the size and power of a given test procedure, you’ll need to know three things:

1. The level of significance (alpha) that you want to use

2. The effect size that you’re interested in detecting

3. The sample size that you have available

With those three pieces of information, you can plug them into the following formulas to get your desired results:

Size:

α=P(reject H0|H0 is true)

Power:

1-β=P(reject H0|H1 is true)

The size of a test is simply the probability of rejecting the null hypothesis when it is true. The power of a test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true.

For example, let’s say you’re interested in conducting a t-test to compare the means of two groups. You want to use an alpha level of 0.05, you’re interested in detecting an effect size of 0.5, and you have a sample size of 100. Plugging those values into the formulas above, we get:

Size:

α=0.05

Power:

1-β=0.95

This means that there’s a 5% chance of incorrectly rejecting the null hypothesis when it is true and a 95% chance of correctly rejecting the null hypothesis when the alternative hypothesis is true.

In general, you want your test to have high power. This means that you want to have a low chance of incorrectly rejecting the null hypothesis when it is true. To increase the power of your test, you can either increase the effect size that you’re interested in detecting or increase the sample size.

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**Assignment Task 6: Obtain a most powerful test of two simple hypotheses using the Neyman Pearson lemma and extend this to a uniformly most powerful test of one-sided alternatives.**

Neyman and Pearson developed the powerful test of two simple hypotheses to more accurately assess the results of epidemiological studies. This test is based on the comparison of two groups, one that is exposed to a potential risk factor (the treatment group) and one that is not (the control group). The Neyman-Pearson test aims to compare the two groups to determine whether there is a statistically significant difference between them.

The Neyman-Pearson test has been used in many studies investigating the effects of treatments and other interventions on health outcomes. In general, this test is considered to be more accurate than other methods, such as the chi-squared test or Fisher’s exact test, because it takes into account the fact that there may be confounding factors that can influence the results of the study.

The Neyman-Pearson test is based on the comparison of two groups, one that is exposed to a potential risk factor (the treatment group) and one that is not (the control group). The Neyman-Pearson test aims to compare the two groups to determine whether there is a statistically significant difference between them.

The Neyman-Pearson test has been used in many studies investigating the effects of treatments and other interventions on health outcomes. In general, this test is considered to be more accurate than other methods, such as the chi-squared test or Fisher’s exact test, because it takes into account the fact that there may be confounding factors that can influence the results of the study.

The Neyman-Pearson lemma is a powerful tool for testing hypotheses and has been used in many different fields, including medicine, finance, and ecology. This lemma allows us to find the most powerful test of two simple hypotheses. To do this, we first need to define what we mean by a “simple hypothesis.” A simple hypothesis can be completely described by a single parameter, such as the mean or the variance.

**Assignment Task 7: Use the likelihood ratio procedure to derive a test of nested hypotheses for some simple statistical models.**

The likelihood ratio is a measure of how much the data favours one model over another. It is defined as the ratio of the maximum likelihoods of the two models and is a non-negative number.

The likelihood ratio can be used to test nested hypotheses. A nested hypothesis is a special case of two competing hypotheses, where one hypothesis is a subset of the other. In other words, the nested hypothesis is just a more specific version of the other hypothesis.

The likelihood ratio can be used to test whether or not the data favours the more specific hypothesis over the less specific one. The test statistic is simply the difference in maximum likelihoods between the two models, divided by their sum. This test statistic has a chi-squared distribution with degrees of freedom equal to the difference in the number of parameters between the two models.

If we are testing whether or not the data favours a model with one additional parameter, then the test statistic has a chi-squared distribution with one degree of freedom.

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**Assignment Task 8: Explain the fundamental concepts of Bayesian statistics and use these concepts to calculate Bayesian estimators.**

Bayesian statistics is a statistical approach that uses past data and prior beliefs to predict future events. Bayesian estimators are calculated using the Bayesian theorem, which states that the probability of an event happening is equal to the product of the prior probability of the event happening and the likelihood of the event happening.

Bayesian estimators are used to calculating the most likely value for an unknown quantity, based on available information. They can be used in many different fields, such as medicine, engineering, and finance. In medicine, Bayesian methods are often used to calculate clinical trial results, diagnostic test accuracy, and treatment effect sizes. In engineering, Bayesian estimators are used to designing optimal experimental setups and estimate point source emission rates. In finance, Bayesian estimators are used to predict stock prices and asset returns.

The prior probability is the probability of an event happening before any new information is taken into account. The likelihood is the probability of an event happening, given that it has been observed.

The posterior probability is the probability of an event happening, after all of the information has been taken into account. The posterior probability is equal to the product of the prior probability and the likelihood.

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