Comparing Two Samples Part 4

Examples of confidence intervals for means One of the main parts of inferential statistics is the development of methods for calculating confidence intervals. Confidence intervals provide us with a method for estimating the population coefficient. Instead of saying that a parameter is equal to a specific value, we say that the parameter falls within a range of values. This range of values is usually an estimate, along with a margin of error that we add and subtract from the estimate. The level of confidence is related to each period. The confidence level gives a measure of how often, in the long run, the method used to obtain our confidence interval captures the true population parameter. When learning about statistics, it is helpful to look at some examples. Below we'll look at several examples of confidence intervals around mean content. We will see that the method we use to create a confidence interval around a mean depends on more information about our population. Specifically, the approach we take depends on whether or not we know the population standard deviation. problem statement We start with a simple random sample of 25 specific newts and measure their tails. The average tail length of our specimen is 5 cm. If we know that 0.2 cm is the standard deviation of the tail lengths of each newt in the community, what is the 90% confidence interval for the average tail length of each newt in the community? If we know that 0.2 cm is the standard deviation of tail lengths for each newt in the community, what is the 95% confidence interval for the average tail length of each newt in the community? If we find that 0.2 cm is the standard deviation of the newt's tail lengths in our sample, what is the 90% confidence interval for the mean tail length for each newt in the population? If we find that 0.2 cm is the standard deviation of newt tail lengths in our population sample, what is the 95% confidence interval for the mean tail length for each newt in the population? discuss problems We begin by analyzing each of these problems. In the first two problems we know the value of the population standard deviation. The difference between these two problems is that the level of confidence is greater in #2 than it is in #1. In the second two problems, the population standard deviation is unknown. For these two problems, we will estimate this parameter using the sample standard deviation. As we saw in the first two problems, here we also have different levels of confidence. Solution We will calculate solutions for each of the above problems. Since we know the population standard deviation, we will use the z-score table. The z-value that corresponds to a 90% confidence interval is 1.645. Using the margin of error formula, we have a confidence interval from 5 - 1.645 (0.2/5) to 5 + 1.645 (0.2/5). (5 in the denominator here because we took the square root of 25). After performing the calculation, we have 4.934 cm to 5.066 cm as the confidence interval for the mean content. Since we know the population standard deviation, we will use the z-score table. The z-value that corresponds to a 95% confidence interval is 1.96. Using the margin of error formula, we have a confidence interval from 5 - 1.96 (0.2/5) to 5 + 1.96 (0.2/5). After performing the calculation, we have 4.922 cm to 5.078 cm as the confidence interval for the mean content. Here we do not know the population standard deviation, only the sample standard deviation. Thus we will use the t-score table. When we use a table of t degrees, we need to know how many degrees of freedom we have. In this case, there are 24 degrees of freedom, which is one degree less than the sample size 25. The t-value corresponding to the 90% confidence interval is 1.71. Using the margin of error formula, we have a confidence interval from 5 - 1.71 (0.2/5) to 5 + 1.71 (0.2/5). After performing the calculation, we have 4.932 cm to 5.068 cm as the confidence interval for the mean content. Here we do not know the population standard deviation, only the sample standard deviation. So we'll use the t-score table again. There are 24 degrees of freedom, which is one degree less than the sample size 25. The t-value corresponding to the 95% confidence interval is 2.06. Using the margin of error formula, we have a confidence interval from 5 - 2.06 (0.2/5) to 5 + 2.06 (0.2/5). After performing the calculation, we have 4.912 cm to 5.082 cm as the confidence interval for the mean content. discuss solutions There are a few things to note when comparing these solutions. The first is that in each case the higher our confidence level, the higher the z or t value we ended up with. The reason for this is that in order to be more confident that we actually averaged the population in our confidence interval, we need a wider time interval. The other advantage to note is that for a given confidence interval, those using t are greater than those with z. The reason for this is that the t distribution has a larger variance in its tails than the standard normal distribution. The key to correct solutions to these types of problems is that if we know the population standard deviation, we use the z-score table. If we do not know the population standard deviation, we use the t-score table. Example of two models of T-test and confidence interval Sometimes in statistics it is useful to see prepared examples of problems. These examples can help us discover similar problems. In this article, we will learn about the process of making inferential statistics of a result relating to two population means. Not only will we see how to perform a hypothesis test about the difference between two population means, but we will also build a confidence interval for that difference. The methods we sometimes use are called a twosample t-test and a two-sample confidence interval t. problem statement Suppose we want to test the mathematical aptitude of elementary school children. One question we might have is whether higher grade levels have higher average test scores. A simple random sample of 27 third-grade students is given a math test, their answers are scored, and the results are found to have a mean score of 75 with a sample standard deviation of 3 points. A simple random sample of 20 fifth graders is given the same math test and their answers are recorded. The mean score for the fifth graders is 84 points with a sample standard deviation of 5 points. Under this scenario, we ask the following questions: Do the sample data provide evidence that the population mean test scores for all fifth graders exceed the population mean test scores for all third graders? What is the 95% confidence interval for the difference in mean test scores between populations of third and fifth graders? Terms and Procedures We must choose which procedure to use. In doing so, we must make sure and verify that the conditions for this procedure are met. We are asked to compare two means of population. One set of methods that can be used to do this are those of the two-sample t-procedures. In order to use these t-measures for two samples, we need to ensure that the following conditions are true: We have two simple random samples from two sets of interest. Our simple random samples make up no more than 5% of the population. The two samples are independent of each other, and there is no match between the participants. The variable is normally distributed. Both the mean and standard deviation are unknown for each of the populations. We see that most of these conditions are satisfied. We were told that we had simple random samples. The population we study is large as there are millions of students at these grade levels. A condition that we cannot automatically assume is whether the test scores are normally distributed. Since we have a large enough sample size, by the robustness of our t procedures, we do not necessarily need the variable to be normally distributed. Since the conditions are met, we perform several preliminary calculations. traditional error The standard error is an estimate of the standard deviation. For this statistic, we add the sample variance to the samples and then take the square root. This gives the formula: ( s 1 2 / n 1 + s 2 2 / n 2 ) 1/2 Using the above values, we see that the standard error value is (3 2 / 27 + 5 2/20 ) 1/2 = (1/3 + 5/4 ) 1/2 = 1.2583 degrees of freedom We can use a conservative approximation of our degrees of freedom. This may reduce the number of degrees of freedom, but it is much easier to calculate than using the Welch formula. We use the smallest sample size, then subtract one from that number. For example, the smaller of the two samples is 20. This means that the number of degrees of freedom is 20-1 = 19. Hypothesis Test We want to test the hypothesis that fifth graders have greater average test scores than third graders. Suppose μ1 is the population average of all 5th graders. Similarly, we allow μ2 to be the population mean scores for all third graders. The hypotheses are as follows: H0:μ1-μ2=0 ha:μ1-μ2>0 The test statistic is the difference between the sample mean, which is then divided by the standard error. Since we use typical standard deviations to estimate the population standard deviation, the test statistic is from a t-distribution. Test statistic value (84 - 75)/1.2583. That's approximately 7.15. We now determine the p-value for this hypothesis test. We look at the value of the test statistic, and where this falls on a t-distribution with 19 degrees of freedom. For this distribution, we have 4.2 x 10 -7 as the p-value. (One way to determine this is to use the T.DIST.RT function in Excel.) Since we have a small p-value, we reject the null hypothesis. The conclusion is that the average test scores for fifth graders are higher than the average test scores for third graders. confidence interval Since we established that there is a difference between the mean scores, we now define a confidence interval for the difference between these two means. We already have a lot we need. The variance confidence interval needs an estimate and a margin of error. It is easy to calculate the estimate of the difference between two means. We simply find the difference in the sample mean. This difference in the sample means an estimate of the difference in the population mean. For our data, the difference in the sample mean is 84-75 = 9. The margin of error is a little trickier to calculate. For this, we need to multiply the appropriate statistic by the standard error. The statistic we need is found by consulting a statistical table or program. Again using conservative approximation, we have 19 degrees of freedom. For the 95% confidence interval, we see that t * = 2.09. We can use the T.INV function in Exce l to calculate this value. Now we add everything together and see that our margin of error is 2.09 x 1.2583, which is approximately 2.63. The confidence interval is 9 ± 2.63. The interval is 6.37 to 11.63 points in the test chosen by fifth and third graders. Calculate the confidence interval for the mean when you know sigma In inferential statistics, one of the main goals is to estimate an unknown population parameter. You start with a statistical sample, and from here you can select a range of values for the parameter. This range of values is called the confidence interval. confidence intervals Confidence periods are similar to each other in a few ways. First, many two-sided confidence intervals have the same shape: Estimate ± margin of error Second, the steps for calculating confidence intervals are very similar, no matter what type of confidence interval you're trying to find. The specific type of confidence interval that will be examined below is a two-sided confidence interval for the population mean when the population standard deviation is known. Also suppose you are working with a normally distributed population. Confidence period for a mean with a known sigma Below is a process for finding the required confidence interval. While all the steps are important, the first one is special: Check the conditions: Start by making sure you meet the conditions of your trust period. Suppose you know the value of the population standard deviation, denoted by the Greek letter sigma σ. Also assume a normal distribution. Estimate calculation: Estimate a population parameter — in this case, the population mean — using a statistic, which in this problem is the sample mean. This involves creating a simple random sample from the population. Sometimes you can assume that your sample is a simple random sample, even if it doesn't fit the strict definition. Critical value: Get the critical value z* that corresponds to your confidence level. These values are found by referring to a z-score table or by using the software. You can use the z-Score table because you know the value of the population standard deviation, and you assume that the population is normally distributed. Common critical values are 1.645 for the 90 percent confidence level, 1.960 for the 95 percent confidence level, and 2.576 for the 99 percent confidence level. Margin of error: Calculate the margin of error z * σ / √ n, where n is the size of the simple random sample you have configured. Conclusion: Finish by combining the estimate and the margin of error. This can be expressed as either an estimate ± margin of error or an estimate - margin of error of the estimate + margin of error. Be sure to clearly state the confidence level associated with your confidence interval. Example To see how you can create a confidence interval, work through an example. Suppose you know that the IQ scores of all freshmen in a college are normally distributed with a standard deviation of 15. You have a simple random sample of 100 freshmen, and the average IQ score for this sample is 120. Find the 90% confidence interval for the mean score IQ of all freshmen in college. Work through the steps above: Check the conditions: The conditions are met since you were told that the population standard deviation is 15 and that you are dealing with a normal distribution. Calculate the estimate: You have been told that you have a simple random sample of size 100. The average IQ of this sample is 120, so this is your estimate. Critical Value: The critical value of the 90 percent confidence level is given by z * = 1.645. Margin of error: Use the margin of error formula and get an error of z * σ / √ n = (1.645) (15) / √ (100) = 2.467. Conclusion: Finish by putting it all together. The 90 percent confidence interval for the population mean IQ is 120 ± 2.467. Alternatively, you could mention this confidence interval from 117.5325 to 122.4675. Practical considerations Confidence intervals of the above type are not very realistic. It is very rare to know the population standard deviation but not the population mean. There are ways in which this unrealistic assumption can be removed. While you were assuming a normal distribution, this assumption need not hold. Nice samples, which do not show any strong deviation or contain outliers, along with a sufficiently large sample size allow you to invoke the central limit theorem. As a result, you are entitled to use the z-score table, even for populations that are not normally distributed.