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Transcript
Confidence Intervals Formulas
Chapter 6
For  (mean): x  z / 2 / n
1. The sample mean, x , estimates the population mean,  and is therefore the center of our interval.
2. The margin of error is made of:
a. z/2 gives us the desired confidence level.
b. the population standard deviation, , which must be known.
c. the sample size, n, which allows us to control the variability of our statistic, x .
3. We must have a random sample so x is unbiased and the standard deviation of x is /n.
4. We are using z-scores, so we must have either a large sample or have sampled from a normal
population.
A confidence interval gives us a ‘range of plausible values for a population parameter’. We are
confident that method here (both the random sampling method and the formula) will produce intervals
which cover the true population parameter (1)*100% of the time.
Sample size determination:
z 
To find the ‘right’ sample size use: n    / 2  , where m is the desired margin of error.
 m 
2
Test Statistics for Hypothesis are just the Z-score formulas:
When testing the mean with  known: H0:  = 10.2 vs. HA:   10.2 The TS: z 
x  10.2
x
n
Chapter 8 (later)
p(1  p)
n
1. The sample proportion, p, estimates the population mean,  and for large enough samples this
interval is sufficient. When the sample is small, a better estimate to use in confidence intervals is the
For  (proportion):
p  z / 2
Wilson estimate, p  x  2 , also called the ‘plus four estimate’.
n4
2. The margin of error is made of:
a. z/2 gives us the desired confidence level.
b. the standard error of p is SE p 
p(1  p) since the standard deviation is dependent on , the parameter
n
we’re having to estimate.
c. the sample size, n, which allows us to control the variability of our statistics, both p and p .
3. We must have a random sample so p(and p ) is unbiased and the standard deviation of p is  (1   ) .
n
4. We are using z-scores, so we must have n and n(1)  10. Since we don’t know , we need np and
n(1p)  10.
When testing the proportion: H0:  = 0.3 vs. HA:   0.3 The TS: z 
p  0.3
0.3(1  0.3)
n