Download Day 9 Confidence Intervals

Confidence Intervals
OBJ: • Determine confidence
intervals for statistical data
Determine the area under the
standard normal curve that lies
between -1 and 1
z
-3 -2 -1 0
1
2
3
Determine the area under the
standard normal curve that lies
between -2 and 2
z
-3 -2 -1 0
1
2
3
Determine the area under the
standard normal curve that lies
between -3 and 3
z
-3 -2 -1 0
1
2
3
The 68% - 95% - 99% Rule:
Every normal distribution has three
important properties:
• About 68% of the distribution is within
1 standard deviation of the mean.
• About 95% of the distribution is within
2 standard deviations of the mean.
• About 99% of the distribution is within
3 standard deviations of the mean.




p (1 – p)
p – 3 s < p < p + 3s √
n
EX: Of 400 households surveyed,
144 reported watching the Channel
7 news at least twice a week. Find a
99% confidence interval for p, the
unknown proportion of households
that watch thee Channel 7 news at
least twice a week.

p = 144
400 = .36


p (1 – p) = .36  .64
√ n
 400
=.024
.36 – 3(
) < p < .36 + 3(
)
.36 – 3(.024) < p < .36 + 3(.024)
.288 < p < .432
28.8% < p < 43.2%




p (1 – p)
p – 2 s < p < p + 2s √
n
EX: Suppose that the general
manager of Channel 5 claims that
less than 30% of the households
watch the Channel 7 news at least
twice a week. Use a 95% confidence
interval to show that it is unlikely
that the manager is correct.

p = 144
400 = .36


p (1 – p) = .36  .64
√ n
 400
=.024
.36 – 2(
) < p < .36 + 2(
)
.36 – 2(.024) < p < .36 + 2(.024)
. 312 < p < .408
31.2% < p < 40.8%
EX: A new, medically
approved wheelchair was
judged to be an
improvement over a
standard wheelchair by
320 out of 400 patients in
a survey.

p = 320
400 = .8


p (1 – p)
√
n
Find a 99% confidence interval
for p, the proportion of the
wheelchair using population for
whom the new wheelchair
would be an improvement.


p (1 – p) = .8  .2
√ n
 400
=.02


p – 3 s < p < p + 3s
.8 – 3( ) < p < .8 + 3( )
.8 – 3(.02)< p <.8 +3(.02)
.74 < p < .86
74% < p < 86%
If 72% of the wheel chair-using
population has found that the
standard wheelchair to be
acceptable, does it appear that
manufacturing the new wheelchair
is a good idea?


p (1 – p)
√
n
EX: a. If 50 out of 100
people surveyed 
favor brand X, find p
and the standard
deviation of p.

p = 50
100 = .5
p (1 – p) = .5  .5
√
n
 100
= .05


p – 2 s < p < p + 2s
Find a 95% confidence
interval of p, the
proportion of the
population favoring
brand X.


p – 2 s < p < p + 2s
.5 – 2(
) < p < .5 + 2(
)
.5 – 2(.05) < p < .5 + 2(.05)
.4 < p < .6
40% < p < 60%