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21.3.2 Size & Power of Test

Confidence Intervals

A point estimate is an estimate of a feature of the population from a sample.

An x% confidence interval for the μ coincides with the acceptance region for a (100 - x)% significance level test for the sample mean.

It does not tell us that the mean lies within the interval with x% confidence. It tells us that if a large number of such intervals were calculated based on different point estimates of μ, we would expect x% of those to contain μ.

lightbulbExample

Example Take the data set

12, 16, 10, 9, 12

If we are given that these are from a normally distributed population, the x\overline{x} is a point estimate for μ and s2s^2 for σ2\sigma^2.

However, due to the nature of statistical error, the point estimate may not be completely accurate; however, we can give bounds within which we know the population parameter we are estimating is likely to lie. These are called confidence intervals.

lightbulbExample

Example A point estimate from a sample of size 50 is calculated to be 26.6 for the mean μ. Given the population is known to have standard deviation σ = 6, construct a 95% confidence interval for μ.

  1. Draw a diagram to aid in determining the correct z-value to calculate:

CALCULATE Z-Value associated with 0.975

  1. The mean μ lies within the confidence interval to the given degree of confidence according to the following formula:
μ=x±Zσn\mu = \overline{x} \pm Z \frac{\sigma}{\sqrt{n}}

Where:

  • x\overline{x} is the center of the interval (sample mean).
  • ZZ is how many standard deviations we want our boundary to lie from x\overline{x}.
  • σn\frac {\sigma}{\sqrt{n}} is the standard deviation of the sampling distribution to reach our Z-value to fit our desired distribution.
26.61.96×650μ26.6+1.96×650\Rightarrow 26.6 - 1.96 \times \frac{6}{\sqrt{50}} \leq \mu \leq 26.6 + 1.96 \times \frac{6}{\sqrt{50}}24.937μ28.263\Rightarrow 24.937 \leq \mu \leq 28.263:success[24.9μ28.3](3sf)\Rightarrow :success[24.9 \leq \mu \leq 28.3] \quad (3sf)

How to Calculate Confidence Intervals on a Graphical Calculator

  • A) Stats Mode

    image
  • B) Interval Mode

    image
  • C) Z Interval

    image
  • D) Select 1-Sample

    image
  • E) Input data:

    image
  • F) Output displays the lower and upper bounds.

    image

Note: "List" mode can be used instead of variable if data is given.

Past Paper Example

infoNote

Q2 (June 2008, Q1)

Some biologists were studying a large group of wading birds. A random sample of 36 were measured, and the wing length x mm of each wading bird was recorded. The results are summarized as follows:

x=:highlight[6046],x2=:highlight[1,016,338]\sum x = :highlight[6046], \quad \sum x^2 = :highlight[1,016,338]

Questions:

(a) Calculate unbiased estimates of the mean and the variance of the wing lengths of these birds.

Given that the standard deviation of the wing lengths of this particular type of bird is actually 5.11 mm.

(Normally we would use s2s^2 to estimate σ2\sigma^2 but to catch us out we have keen gives the population σ

(b) Find a 99% confidence interval for the mean wing length of the birds from this group.


Solution: (a)

μx=604636=:success[167.9444]\mu \approx \overline{x} = \frac{6046}{36} = :success[167.9444]σ2=s2=(x2x2)×nn1\sigma^2 = s^2 = \left( \overline{x^2}-\overline{x}^2 \right) \times \frac{n}{n-1}=(1,016,33836167.94442)×3635=:success[29.179]= \left( \frac{1,016,338}{36} - 167.9444^2 \right) \times \frac{36}{35}= :success[29.179]

Solution: (b)

99%: z=:highlight[2.5758]\quad z = :highlight[2.5758] \quad (from the z-table)

CI (Confidence Interval) Calculation:

CI = x±zσn\overline{x} \pm z \frac{\sigma}{\sqrt{n}}

Substituting values:

167.944±2.5758×5.136167.944 \pm 2.5758 \times \frac{5.1}{\sqrt{36}}

CI = (165.75, 170.13) \quad \color{green}\checkmark

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