Transcript 3) Calculate the interval

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
2) The sample distribution should be approximately normal
- the population is known to be normal, or
- the sample size is large (n  30) (CLT), or
- graph data to show approximately normal
3) 10% rule – The sample should be less than 10% of the
population
4) σ is known (or unknown)
3) Calculate the interval
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If σ is known we perform a z-interval
If σ is unknown we perform a t-interval
4) Write a statement about the interval in the context
of the problem.
Confidence interval for a population
mean (z-interval):
Standard
Critical
value
deviation of the
statistic
  
x  z *

 n
estimate
Margin of error
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 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 = the true mean systolic blood pressure of healthy white males
Justify the confidence interval needed (state assumptions)
1) The sample must be random which is stated in the problem.
2) The sample distribution should be approximately normal.
Since n = 54 >30, by the CLT we can assume the sample
distribution is approximately normal.
3) The sample should be less than 10% of the population.
The population should be at least 540 healthy white males,
which I will assume.
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
95% CI
df  53
 9.3 
114.9  2.009 

 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.
Steps for a hypothesis test :
1) Define the parameter
2) Hypothesis statements
3) Assumptions
4) Calculations (Find the p-value)
5) Decision and Conclusion,
Conditions for one-sample means
1) The sample should be chosen randomly
2) The sample distribution should be approximately normal
- the population is known to be normal, or
- the sample size is large (n  30) (CLT), or
- graph data to show approximately normal (normal
probability plot and box plot)
3) 10% rule – The sample should be less than 10% of the
population
4) σ is known or unknown
If σ is known we perform a z-test
If σ is unknown we perform t-test
Formulas:
 known:
statistic - parameter
test statistic 
standard deviation of statistic
z=
x  x

n
Formulas:
 unknown:
statistic - parameter
test statistic 
standard deviation of statistic
t=
x
s
n
x
df  n  1
Example 2 The Fritzi Cheese Company buys milk from
several suppliers as the essential raw material for its
cheese. Fritzi suspects that some producers are
adding water to their milk to increase their profits.
Excess water can be detected by determining the
freezing point of milk. The freezing temperature of
natural milk varies normally, with a mean of -0.545
degrees and a standard deviation of 0.008. Added
water raises the freezing temperature toward 0
degrees, the freezing point of water (in Celsius). The
laboratory manager measures the freezing
temperature of five randomly selected lots of milk
from one producer with a mean of -0.538 degrees. Is
there sufficient evidence to suggest that this
producer is adding water to his milk?
Parameters and Hypotheses
μ = the true mean freezing temperature of milk
H0: μ = -0.545
Ha: μ > -0.545
Assumptions (Conditions)
1) The sample must be random which is stated in the problem.
2) The sample distribution should be approximately normal. Since it is
stated in the problem that the population is normal then the sample
distribution is normal.
3) The sample should be less than 10% of the population. The population
should be at least 50 lots of milk, which I will assume.
4)  is known
Since the conditions are met, a z-test for the one-sample means is appropriate.
Calculations
x    .545
x  .538
n5
  .008
 = 0.05
z
x  x

n
.538  (.545)
 1.9565

.008
5
p  value  P( z  1.9565)  .0252
.0252  .05
Decision: Since p-value < , I reject the null hypothesis at the .05
level.
Conclusion:
There is sufficient evidence to suggest that the true mean freezing
temperature is greater than -0.545. This suggests that the
producer is adding water to the milk.
Example 3 (page 545 #33)
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
(no kidding) purchased some randomly selected bags of
cookies, and counted the chocolate chips. Some of their
data are give 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.
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 sample must be random which is stated in the problem.
2) The sample distribution should be approximately normal.
The normal probability plot is fairly linear and the boxplot shows no
outliers, so we will assume that the sample distribution is approximately
normal.
3) The sample should be less than 10% of the population. The population
should be at least 160 bags of Chips Ahoy, which we will assume.
4)  is unknown
Since the conditions are met, a t-test for the one-sample means is appropriate.
Calculations
x  x
 x    1000 t  s 
n
x  1238.1875
n  16
s  94.282
 = 0.05
df  15
 (1000)
1238.1875
94.282
 10.1053
16
p  value  P(t  10.1053)  2.176 108
8
2.176 10  .05
Decision: Since p-value < , I reject the null hypothesis at the .05
level.
Conclusion:
There is sufficient evidence to suggest that the true mean number
of chocolate chips in each bag of Chips Ahoy is greater than 1000.