# sections 7.1b-7.3

```Example:
Suppose you administer a certain aptitude test to a random sample of 36 students
in your school, and that the average score is 105. We want to determine the mean
of the population of all students in the school. Assume a standard deviation of
σ = 15 for the test.
What is the z* value for a 98% confidence interval?
Sketch the distribution of sample means. Label the x-axis appropriately out to 3
standard deviations.
Determine the 98% confidence interval for the mean score
school.
µ for the whole
What is the margin of error?
Write a sentence that explains the significance of the confidence interval.
What sample size would be needed to have a margin of error at most 4 points?
How would a confidence interval of this same data change if the confidence level
Section 7.2 - Large-Sample CI’s for Population Mean and Proportion
The assumption that we have made is that we know the standard deviation of the
population. If we don’t know the standard deviation of the distribution, we may
use the sample standard deviation,
s 2 as a point estimate for σ
2
.
Example: The mean breaking strength x of a sample of n = 112 steel beams is
42,196 pounds per square inch, and the sample standard deviation is s = 614 .
Provide a 90% confidence interval for the mean breaking strength µ. Provide a
95% confidence interval for µ.
Rule of Thumb: We can use s to approximate σ when n > 30.
Proposition:
If n is sufficiently large, a (1 – α) confidence interval for µ is
s
x ± zα /2
n
This can be used to give an approximate confidence interval regardless of the
distribution of the population.
Confidence Intervals for Proportions
Suppose we have a population in which each member either has a trait or does not
have the trait. We may obtain a confidence interval for the proportion of the
population with the trait, p.
If a sample of size n is taken from our population, and y of the sample have the
trait, then an approximate (1− α )100% confidence interval for the proportion p of
the population with the trait is
pˆ ± zα / 2
ˆˆ
pq
n
y
where pˆ = and qˆ = 1 − pˆ
n
This interval is sufficient if n is large, the actual confidence interval appears in
Choosing the sample size:
z
*
p * (1 − p * )
≤m
n
Use p*=.5 if you don’t have a better guess.
We will use the simpler version, but keep in mind that there are other methods.
Example: Suppose we interview 200 voters and 104 say they plan to vote for a
certain candidate. Determine a 90% confidence interval for the proportion of the
population that plan to vote for the candidate.
Example: An experimenter flips a coin 100 times and gets 54 heads.
Find an approximate 90% confidence interval for the probability of flipping a head
with this coin.
Suppose that prior to conducting the coin-flipping experiment, we suspect that the
coin is fair. How many times would we have to flip the coin in order to obtain a
90% confidence interval of width of at most 0.1 for the probability of flipping a
EMCF 19
1. In an opinion poll, 25% of 200 people sampled said they were strongly
opposed to the state lottery. The standard error of the sample proportion is
approximately
a. .03
b. .25
c. .00094 d. 6.12
e. .06
A bag of M&Ms was randomly selected from the grocery store shelf, and the color
counts were:
Brown 16 Red 11 Yellow 19
Orange 5 Green 7 Blue
3
2. Construct a 98% confidence interval for the probability of a brown M&M.
a. (.20, .32)
b. (.13, .39)
c. (.37, .63)
d. (.13, .26)
e. none of these
Section 7.3 - Intervals Based on a Normal Population Distribution (Small
Samples)
The t distribution:
Let x1 , x2 ,..., xn constitute a random sample from a normal population
distribution. Then the probability distribution of the standardized variable
x −µ
t=
s
n
is the t distribution with ( n − 1) degrees of freedom.
Confidence Interval (small n or σ unknown)
When x is the sample mean from a population whose distribution is normal (or at
least approximately normal), n is small, and σ is unknown, the (1− α ) confidence
interval for µ is the interval
x ± tα / 2
s
n
Example:
For a t distribution with 8 degrees of freedom, find c so that P (T > c ) = 0.025
If T has 25 degrees of freedom, what would c be?
Example: The mean µ of the times it takes a lab technician to perform a certain
task is of interest. If the lab technician was observed on n = 16 different occasions
and the mean and standard deviation of these times were 4.3 minutes and 0.6
minutes respectively. Give a 95% confidence interval for her mean completion
time µ.
EMCF 19
3. A
4. The weight of 9 men have mean 175 lbs. and a standard deviation 15 lbs. What
is the standard error of the mean?
a. 58.3
b. 19.4
c. 5
d. 1.7
5. What is the critical value t* which satisfies the condition that the t distribution
with 8 degrees of freedom has probability 0.10 to the right of t*?
a. 1.397
b. 1.282
c. 2.89
d. 0.90
```