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Confidence Intervals
Means
Remember: In sampling
distribution of means…..

x  
 x 


n
Approximately normal if
is large ( n  30) – CLT
 The population is normal
n
So how will I find a confidence intervals for a
mean? How is it different from a proportion?
Formula for a z-confidence interval:
  
x  z 

2
 n
Suppose you work for a consumer advocate agency & want to find the
mean repair cost of a washing machine. You randomly select 40 repair
costs and find the mean to be $100. The standard deviation is $17.50.
construct a 95% interval for the population mean.
Find the minimum required sample size if you want to be 95% confident
that the sample mean is within 2 units of the population mean if the
population standard deviation is 4.8.
But what if the sample size is small or the
population standard deviation is unknown?
We use the t-distribution if the population
standard deviation is unknown.
It’s bell shaped – centered at 0.
 Each t-distribution is more spread out than
the z-distribution (normal).
 As the sample size increases, the spread
decreases and actually approaches the
normal distribution.
 Based on degrees of freedom (df = n-1)

Degrees of freedom: (n-1)

Suppose A + B + C + D = 18
Free to be
anything

Thus there are 3 degrees of freedom
The t- distribution compared to the
normal distribution

http://www.nku.edu/~longa/stats/taryk/TDist
.html

In normal sampling but with t-distribution
z
x

n
x
t
s
n
Reading the
t-chart
The Raman Arches is an Italian restaurant. The manager wants to
estimate the average amount a customer spends on lunch. A random
sample of 115 customers’ lunch tabs gave a mean of $9.74 with a
standard deviation of $2.93. Find and interpret a 90% confidence interval
for the average amount spent on lunch by all customers.
A sample of 20 students had a test average of 100 with a
standard deviation of 4.2 points. Find and interpret a 90%
confidence interval for the average test score for all students
on this test.
A random sample of large tents listed in Consumer Reports: Special
Outdoor Issue gave the following prices. Find a 90% confidence
interval for the mean price of all such tents.
115 110
140 135
80
110
150 210
250 120
230 130
Homework - Worksheet