Estimating the Value of Parameters
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| What do we mean by a point estimate? | A point estimate is the value of a STATISTIC that estimates the value of a PARAMETER.
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| 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”.
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| 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.
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| 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%.
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| 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.
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| 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.
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| 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.
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| 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:
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| 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.
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| 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.
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| 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.
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| 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.
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| 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.
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| 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.
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| 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.
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| 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.”
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| 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)]
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| 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)]
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| 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]
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| 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.
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| 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.
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| 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”.
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| 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.
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| 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.
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| 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)]
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| 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.
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| 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.
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| 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.
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| 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.”
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| 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).
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| 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%.
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