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Ch9_Conf. Intervals

Estimating the Value of Parameters

QuestionAnswer
What do we mean by a point estimate? A point estimate is the value of a STATISTIC that estimates the value of a PARAMETER.
What would be a reasonable point estimate for the parameter µ? A reasonable point estimate for the parameter µ would be the statistics “x-bar”.
How is a confidence interval difference from a point estimate? A point estimate uses ONE number to estimate an unknown parameter; a confidence interval uses an INTERVAL of numbers to estimate an unknown parameter.
In constructing confidence intervals, how should we interpret a “level of confidence”? The level of confidence represents the expected proportion of intervals that will contain the parameter if a large number of different samples is obtained. The level of confidence is denoted (1 – α)•100%.
How should we interpret a 95% level of confidence? For example, a 95% level of confidence (α = 0.05), implies that if 100 different confidence intervals are constructed, we will expect 95 of the intervals to contain the parameter and 5 to not include the parameter.
In general, what is the form of a confidence interval estimate for a parameter? Confidence interval estimates for a population parameter are of the general form: Point estimate ± margin of error.
What is the “margin of error” of a confidence interval estimate intended to represent? The margin of error of a confidence interval estimate for a parameter is intended to measure the accuracy of the point estimate.
What are the three factors that determine the size of the margin of error? The size of the margin of error depends on 1) the Level of confidence, 2) the Sample size, and, 3) Standard deviation of the population:
If we keep the sample size the same, but want to increase the level of confidence, what happens to the margin of error? If we keep the sample size the same and increase the level of confidence, the margin of error would also have to increase.
What happens to the margin of error if we keep the level of confidence the same, but increase the sample size? As the size of the random sample increases, assuming we keep the level of confidence the same, the margin of error decreases.
How does the population standard deviation affect the margin error for the confidence interval? The larger the population standard deviation, the larger the margin of error will be for a given level of confidence.
If we want to estimate the population mean, µ, using a confidence interval, what criteria must be met first? Before we can construct a confidence interval for a population mean, µ, we must: 1) know that the population is normally distributed; OR, 2) the sample size n ≥ 30.
If we don’t know the distribution of the population, and the sample size is small, what must we do before it is okay to construct a confidence interval for a population mean, µ? If the sample size is small (i.e., n < 30), in order to construct a confidence interval for a population mean, µ, we must know: 1) whether it is reasonable to assume the data come from a normal population; AND, 2) There are no outliers in the sample.
How to we determine whether it is reasonable to assume the population is normal? To determine whether it is reasonable to assume the population is normal, we would construct a Normal Probability Plot. If MINITAB is used, to be normal all the points must be in the boundary lines.
How to we determine that there are no outliers? To check for outliers, we would construct a boxplot. The boxplot identifies potential outliers as isolated an isolated point or isolated points beyond the whiskers of the graph.
How would we construct a confidence interval for a population mean, µ, if the population standard deviation, σ, is known? The confidence interval would be constructed as: x-bar ± z(α/2) ∙ [σ/sqrt(n)] We will call this a “Z-Interval.”
For a 95% level of confidence, what would the formula for constructing a confidence interval for a population mean, µ, look like? For a 95% level of confidence , the confidence interval would be constructed as: x-bar ± 1.96 ∙ [σ/sqrt(n)]
For a 90% level of confidence, what would the formula for constructing a confidence interval for a population mean, µ, look like? For a 90% level of confidence , the confidence interval would be constructed as: x-bar ± 1.645 ∙ [σ/sqrt(n)]
Suppose we find a 99% confidence interval for a population mean, µ, to be: (2.452, 2.476). How would we interpret this result? "Interpretation": We are 99% confident that the true population mean, µ, is between 2.452 and 2.476. [WE DO NOT SAY IT IS A 99% PROBABILITY THAT THE MEAN IS BETWEEN 2.452 AND 2.476]
Compare the characteristics of the t-distribution to the characteristics of the Standard Normal Distribution (Z-distribution) Both are symmetric and with a mean of zero. The t-distribution does NOT have standard deviation of 1 but tends to be wider than the z-distribution. As the sample size increases, the t-distribution becomes more like the z-distribution.
When would we do not know the population standard deviation, σ, how would we construct a confidence interval for a population mean, µ? When we do not know the population standard deviation, σ, we would construct a confidence interval based a “t-critical value”, rather than a “z-critical value” and we would substitute the sample standard deviation, s, for σ.
How would the confidence interval for a population mean, µ look if the population standard deviation, σ, is not known? The confidence interval for a population mean, µ, if the population standard deviation, σ, is not known would be constructed as: x-bar ± t(α/2) ∙ [s/sqrt(n)] We will call this a “T-Interval”.
When finding t(α/2), what factors are used to calculate the value? To calculate t(α/2), we would need to know 1) the level of confidence, and 2) the degrees of freedom. Recall, the degrees of freedom is “n – 1”, where “n” is the sample size.
For a 95% confidence level, with as sample size of 20, what is t(α/2)? Using a 95% confidence interval, α = 0.05; and for a sample size of 20, degrees of freedom = 19. Therefore, we want to find t(0.025) with 19 degrees of freedom. So, here t(0.025) = 2.093, so – t(0.025) = – 2.093.
So, for a 95% level of confidence where the sample size n = 20, what would the formula for constructing a confidence interval for a population mean, µ, look like, with σ unknown? The 95% confidence interval would be constructed as: x-bar ± 2.093 ∙ [s/sqrt(n)]
What requirement(s) must be met before we can construct a T-interval for a population mean? The requirements are the same whether you are constructing a Z-Interval or a T-Interval for a population mean, µ. That is, we must: 1) know that the population is normally distributed; OR, 2) the sample size n ≥ 30.
In constructing a confidence interval to estimate a population proportion, p, what would we use for the point estimate? The point estimate for the population proportion is the sample proportion, “p-hat”: p-hat = x/n where x is the number of individuals in the sample with the specified characteristic and n is the sample size.
What requirements must be met before we can construct a confidence interval for a population proportion? There two main requirements: 1) The sample size, n, must be 5% of less of population size (n ≤ 0.05 N). This to ensure INDEPENDENCE; 2) To ensure the distribution of the p-hats is NORMAL, we need np(1 – p) ≥ 10.
How would we construct a confidence interval for a population proportion, p? The confidence interval would be constructed as: p-hat ± z(α/2) ∙ sqrt[p-hat(1 – p-hat)/n] We will call this a “1-PropZ-Interval.”
Suppose in a sample of size 1783 there are 1123 individuals with the specified characteristic. Construct a 90% confidence interval for the population proportion. The 90% confidence interval would be given as: (0.611, 0.649).
Interpret the confidence interval, (0.611, 0.649), you used calculated for the population proportion, “p”. We are 90% confident that the true population proportion of individuals with the specified characteristic is between 61.1% and 64.9%.
Created by: wgriffin410
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