Transcript File - phsapstatistics

WARM – UP Quiz Review
An insurance company checks police records on 582
accidents selected at random. Teenagers were involved
in 91 of them.
a.) Find the 95% Confidence Interval for the TRUE % of
accidents that involved teens. (Be sure to interpret
your interval and check the assumptions!) What does
p represent?
a.)
An insurance company checks police records on 582
accidents selected at random. Teenagers were involved in
91 of them. Find the 95% Confidence Interval for the % of
accidents that involved teens. (Follow all steps)
pˆ 1  pˆ 
pˆ  z 
n

One Proportion
0.1564 1-0.1564 
0.1564  1.960 
Z– Conf. Int.
582
We can be 95% confident that the true proportion of
accidents involving teenage drivers is between 0.127
and 0.186.
1. SRS – Stated
2. Appr. Normal: 582(.1564) ≥ 10 582(1 - .1564) ≥ 10
91.025
490.975
3. Population of National Auto Accidents ≥ 10(582)
Review
EXAMPLE #1: An SRS of 500 students leads to a result of 12
National Merit Students.
1.Draw the Sampling Distribution.
SE =
pˆ (1- pˆ )
n
.024 1  .024 
 .007
500
2. Check the Assumptions.
1. SRS – Stated
2. Approx. Norm: 500(.024)=12 ≥ 10
500(1 - .024)=488 ≥ 10
3. Population of Students ≥ 10(500)
3.Find the Probability that 12 or more of 500 students would
be National Merits given that the true proportion = 0.01.


P  pˆ  .024   ?


P  z  3.146   ?
normalcdf (3.146, E 99)  0.0008
P



z
.024  .01 
?
.011  .01 

500

Review
EXAMPLE #2: An SRS of 50 students leads to an average
SAT score of 2100 with σ = 100.
1.Draw the Sampling Distribution.
s
100
SE =
=
 14.142
n
50
2.Check the Assumptions.
1. SRS – Stated
2. Appr. Normal due to Large n and C.L.T.
3. Find the Probability that 50 students will have an average
score greater than 2100 if the true mean SAT score = 2000.


P  x  2100   ?
2100  2000
P  z  7.0711  ?
normalcdf (7.071, E 99)  0
P 


z
100
50
?



Quiz Review
EXAMPLE #3:
A national study was interested in the proportion of high
school students that take AP Statistics. An SRS of 196
students are selected where 20 indicated they are in Stats.
a.) Find the margin of error for a 95% Confidence Interval
m= z
*
pˆ (1- pˆ )
n
.102 (1- .102)
= 0.042
= 1.960
196
Quiz Review
EXAMPLE #4:
-Central Limit Theorem
A Large Random Sample will produce an
approximately normal distribution, regardless of
Population.
-z score for proportion
-z score for means
pˆ  z
*
pˆ 1  pˆ 
n
What happens to the width of the Confidence interval if
you…
1. Increase your Confidence Level?
Interval gets wider
2. Increase your sample size?
Interval gets Narrower.
The Difference between Confidence Level & Intervals
Vitamin D from natural sunlight is important for strong healthy
bones. But with children increasingly staying inside to play
Video games the risk of Vitamin D deficiency is on the rise. A
recent study of 2700 random children found 20% deficient.
1.) Interpret what a 98% Level means.
2.) Find and Interpret the 98% Confidence Interval.
1.) A 98% Confidence Level means that in Repeated
Sampling, 98% of all random samples will produce
intervals that contain the true proportion, p.
The Difference between Confidence Level & Intervals
Vitamin D from natural sunlight is important for strong healthy
bones. But with children increasingly staying inside to play
Video games the risk of Vitamin D deficiency is on the rise. A
recent study of 2700 random children found 20% deficient.
2.) Find and Interpret the 98% Confidence Interval.
2.) p = The true proportion of children deficient in Vitamin D.
1. SRS – Stated
2. Population of Children ≥ 10(2700)
3. 2700(.20) ≥ 10
2700(1 - .20) ≥ 10
pˆ 1  pˆ 
pˆ  z 
n

One Proportion
Z– Conf. Int.
We can be 98% confident that the true proportion of
Vitamin D deficient children is between 18.2% and
21.8%.
WARM – UP Quiz Review
1. The heights of people in a certain population are
normally distributed with a mean of 72 in and a standard
deviation of 2.5 in. An SRS of 36 people are selected
from the population.
a.) Describe the Distribution.
b.)
Verify the conditions are met.
c.)
What is the probability that the sample will yield an
average height less than 68 inches?
EXAMPLE: Vitamin D from natural sunlight is important for
strong healthy bones. But with children increasingly staying
inside to play Video games the risk of Vitamin D deficiency is
on the rise. A recent study of 2700 random children found
20% deficient. 1.) Interpret what a 98% Level means.
2.) Find and Interpret the 98% Confidence Interval.
Check the Assumptions/Conditions…..
1. SRS – Stated
2. Population of children ≥ 10(2700)
3. 2700(.20) ≥ 10
2700(1 - .20) ≥ 10