Download CH 22 Inference for means

Chapter 22
Inference for OneSample Means
Steps for doing a confidence interval:
1)
2)
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State the parameter
Conditions
1) The sample should be chosen randomly or there is random
assignment to treatments
2) The sample distribution should be approximately normal
- the population is known to be normal, or
- the sample size is large (n  30), or
- graph data to show approximately normal
3) 10% rule – The sample should be less than 10% of the
population
4) σ will almost always be unknown
3) Calculate the interval
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If σ is unknown we perform a t-interval…the t distribution
is based on an unknown standard deviation and different
sample sizes (known as degrees of freedom)
4) Write a statement about the interval in the context
of the problem.
Formula for a t-confidence interval:
Critical value
 s 
x  t *

 n
estimate
Margin of error
Standard
deviation of
statistic
df  n  1
Degrees of
freedom
In a randomized comparative experiment
on the effects of calcium on blood
pressure, researchers divided 54 healthy,
white males at random into two groups,
takes calcium or placebo. The paper
reports a mean seated systolic blood
pressure of 114.9 mmHg with standard
deviation of 9.3 for the placebo group.
Find a 95% confidence interval for the
true mean systolic blood pressure of the
placebo group.
State the parameters
μ = population mean systolic blood pressure of healthy white males
Justify the confidence interval needed (state assumptions)
1) The 54 white males were randomly assigned to either the
calcium group or the placebo group.
2) The sample distribution should be approximately normal.
Since n = 54 >30, by the Central Limit Theorem we can
assume the sample distribution is approximately normal.
3) We will assume that this group of white males is
representative of the larger population of white males.
4)  is unknown
Since the conditions are satisfied a t – interval for means
is appropriate.
Calculate the confidence interval.
x  114.9
 s
xt
n  54
n
s  9.3
 9.3 
95% CI
df  53

114.9  2.006
 54 
112.36,117.44
Explain the interval in the context of the problem.
We are 95% confident that the true mean systolic
blood pressure for healthy white males is between
112.36 and 117.44 mmHg.
Conditions for one-sample means
1) The sample should be chosen randomly or there is random
assignment to treatment groups
2) The sample distribution should be approximately normal
- the population is known to be normal, or
- the sample size is large (n  30), or
- graph data to show approximately normal
3) 10% rule – The sample should be less than 10% of the
population
4) σ will almost certainly be unknown, which is why we
conduct t-test for means.
Formula:
 unknown:
statistic - parameter
test statistic 
standard deviation of statistic
t=
x
s
n
x
df  n  1
Example In 1998, as an advertising campaign, the
Nabisco Company announced a “1000 Chips Challenge,”
claiming that every 19-ounce bag of their Chips Ahoy
cookies contained at least 1000 chocolate chips.
Dedicated Statistics students at the Air Force Academy
purchased some randomly selected bags of cookies, and
counted the chocolate chips (no kidding) . Some of their
data are given below.
1219 1214 1087 1200 1419 1121 1325 1345
1244 1258 1356 1132 1191 1270 1295 1135
What does this say about Nabisco’s claim? Test an
appropriate hypothesis at 5% significance.
Parameters and Hypotheses
μ = the true mean number of chocolate chips in each bag of
Chips Ahoy
H0: μ = 1000
Ha: μ > 1000
Assumptions (Conditions)
1) The 16 bags of Chips Ahoy were randomly sampled.
2) The sample distribution should be approximately normal. (Check with an
appropriate graphical display, usually a histogram or boxplot)
The boxplot shows no outliers and is roughly symmetric, so assume
that the sample distribution is approx. normal.
3) The sample should be less than 10% of the population, which would be
at least 160 bags of Chips Ahoy. This is a very safe assumption.
4)  is unknown
Since the conditions are met, a t-test for the one-sample means is appropriate.
Calculations
x  1238.188
n  16
 = 0.05
df  15
x   x 1238.188  1000
t

 10.105
94.282
 s 


15
 n
p  value  P ( t  10.1053)  2.18  108
2.176  108  .05
Decision and conclusion:
Since p-value of .0000000218 is much less than the
significance level of .05, I reject the null hypothesis.
There is overwhelming evidence to suggest that the true
mean number of chocolate chips per bag of Chips Ahoy
is greater than 1000.