AP Statistics Curriculum 2007 Estim S Mean

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Revision as of 05:00, 26 October 2009

Contents

General Advance-Placement (AP) Statistics Curriculum - Estimating a Population Mean: Small Samples

Now we discuss estimation when the sample-sizes are small, typically < 30 observations. Naturally, the point estimates are less precise and the interval estimates produce wider intervals, compared to the case of large-samples.

Point Estimation of a Population Mean

The population mean may also be estimated by the sample average using a small sample. That is the sample average \overline{X_n}={1\over n}\sum_{i=1}^n{X_i}, constructed from a random sample of the process {X_1, X_2, X_3, \cdots , X_n}, is an unbiased estimate of the population mean μ, if it exists! Note that the sample average may be susceptible to outliers.

Interval Estimation of a Population Mean

For small samples, interval estimation of the population means (or Confidence Intervals) are constructed as follows. Choose a confidence level (1 − α)100%, where α is small (e.g., 0.1, 0.05, 0.025, 0.01, 0.001, etc.). Then a (1 − α)100% confidence interval for μ is defined in terms of the T-distribution:

CI(\alpha): \overline{x} \pm t_{\alpha\over 2} E.
  • The Margin of Error E is defined as

E = \begin{cases}{\sigma\over\sqrt{n}},& \texttt{for-known}-\sigma,\\
{SE},& \texttt{for-unknown}-\sigma.\end{cases}

  • The Standard Error of the estimate \overline {x} is obtained by replacing the unknown population standard deviation by the sample standard deviation:

SE(\overline {x}) = {1\over \sqrt{n}} \sqrt{\sum_{i=1}^n{(x_i-\overline{x})^2\over n-1}}

Assumptions

To use the small-sample method for constructing a confidence interval for the population mean, we must have evidence that the data we observed and are using for the point and interval estimates come from a distribution which is (approximately) Normal. If this assumption is violated, the interval estimate we compute may be significantly misrepresenting the real confidence interval!

Example

To parallel the example in the large sample case, we consider again the number of sentences per advertisement as a measure of readability for magazine advertisements. A random sample of the number of sentences found in 10 magazine advertisements is listed. Use this sample to find point estimate for the population mean μ (sample-mean=22.1, sample-variance=737.88 and sample-SD=27.16390579).

16 9 14 11 17 12 99 18 13 12

A confidence interval estimate of μ is a range of values used to estimate a population parameter (interval estimates are normally used more than point estimates because it is very unlikely that the sample mean would match exactly with the population mean) The interval estimate uses a margin of error about the point estimate.

Before you find an interval estimate, you should first determine how confident you want to be that your interval estimate contains the population mean.

80% confidence (0.80), {\alpha \over 2}=0.1, t(df=9) = 1.383
90% confidence (0.90), {\alpha \over 2}=0.05, t(df=9) = 1.833
95% confidence (0.95), {\alpha \over 2}=0.025, t(df=9) = 2.262
99% confidence (0.99), {\alpha \over 2}=0.005, t(df=9) = 3.250

Notice that for a fixed α, the t-critical values (for any degree-of-freedom) exceeds the corresponding normal z-critical values, which are used in the large-sample interval estimation.

Known Variance

Suppose that we know the variance for the number of sentences per advertisement example above is known to be 256 (so the population standard deviation is σ = 16).

  • For {\alpha \over 2}=0.1, the 80%CI(μ) is constructed by:
\overline{x}\pm 1.383{16\over \sqrt{10}}=22.1 \pm 1.28{16\over \sqrt{10}}=[15.10 ; 29.10]
  • For {\alpha \over 2}=0.05, the 90%CI(μ) is constructed by:
\overline{x}\pm 1.833{16\over \sqrt{10}}=22.1 \pm 1.833{16\over \sqrt{10}}=[12.83 ; 31.37]
  • For {\alpha \over 2}=0.005, the 99%CI(μ) is constructed by:
\overline{x}\pm 3.250{16\over \sqrt{10}}=22.1 \pm 3.250{16\over \sqrt{10}}=[5.66 ; 38.54]

Notice the increase of the CI's (directly related to the decrease of α) reflecting our choice for higher confidence.

Unknown Variance

Suppose that we do not know the variance for the number of sentences per advertisement but use the sample variance 737.88 as an estimate (so the sample standard deviation is s=\hat{\sigma}=27.16390579).

  • For {\alpha \over 2}=0.1, the 80%CI(μ) is constructed by:
\overline{x}\pm 1.383{27.16390579\over \sqrt{10}}=22.1 \pm 1.383{27.16390579\over \sqrt{10}}=[10.22 ; 33.98]
  • For {\alpha \over 2}=0.05, the 90%CI(μ) is constructed by:
\overline{x}\pm 1.833{27.16390579\over \sqrt{10}}=22.1 \pm 1.833{27.16390579\over \sqrt{10}}=[6.35 ; 37.85]
  • For {\alpha \over 2}=0.005, the 99%CI(μ) is constructed by:
\overline{x}\pm 3.250{27.16390579 \over \sqrt{10}}=22.1 \pm 3.250{27.16390579\over \sqrt{10}}=[-5.82 ; 50.02]

Notice the increase of the CI's (directly related to the decrease of α) reflecting our choice for higher confidence.

Hands-on Activities

  • See the SOCR Confidence Interval Experiment.
  • Sample statistics, like the sample-mean and the sample-variance, may be easily obtained using SOCR Charts. The images below illustrate this functionality (based on the Bar-Chart and Index-Chart) using the 30 observations of the number of sentences per advertisement, reported above.

Problems




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