Download Math 4, Unit 1, Central Limit Thm/Confidence Intervals

Math 4, Unit 1, Central Limit Thm/Confidence Intervals
Wksht. 8.05 – Confidence Intervals
Name: _________________________
Date: _____________
Confidence Interval & Confidence Level:
We all “know” that normal body temperature is 98.6ºF. Some researchers, Wasserman,
Mackowiak, and Levine, at the University of Maryland, suspected that 98.6ºF might be a little
high. Once a mean of a sample is found, it will represent the best estimate for the population’s
mean; it does not, however, give us any indication of just how good the best estimate is.
Statisticians developed the confidence interval. A confidence interval is a range (i.e. interval)
of values used to estimate the true value of a population parameter. A confidence interval is
associated with a confidence level, such as 0.95 (or 95%). The confidence level gives us the
success rate of the procedure used to construct the confidence
interval. The confidence level is often
expressed as the probability (or area) 1 – ,
where is the complement of the
confidence level. For example, for a 0.95
1-
(95%) confidence level,  = 0.05. A visual
is given to the right.
Margin or Error:
When we collect a sample of data, such as the set of
50 body temperatures to the right, we can calculate
the sample mean, x, and that sample mean is
typically different from the population mean, .
The difference between the sample mean and the
population mean is an error. We recently learned
that

n
Using
is the standard deviation of sample means.

and z notation, we now use the
2
n
following equation to express the margin of error
E: E  z
z


2

2
Level of

z
1–
2
90%
5%
1.645
95%
2.5%
1.96
98%
1%
2.33
99%
0.5%
2.575
2
n
is simply the z-score associated with
2
2
the confidence level. For example, if the
confidence level is 95%, there is 2.5% at
95%
the lower end of the normal curve and
2.5% at the upper end. Since z-scores deal
with area to the left of a point on the
2.5%
2.5%
normal distribution, we are looking for the
z-score that yields a 0.975. That score is
1.96. The confidence interval is simply x ± E.
EXAMPLE 1: For the body temperatures to the right, n = ______ and x = ________ (nearest
100th). Assume that the sample is a simple random sample and that  is
somehow known to be 0.62ºF. Using a 95% confidence interval, find a) the
margin of error and b) the confidence interval for .
98.6
98.6
98.0
97.3
97.2
98.6
98.8
98.3
97.6
98.4
98.0
98.6
98.5
98.2
98.6
98.0
97.0
97.3
99.6
98.2
99.0
97.0
98.7
98.7
98.0
98.4
98.8
97.4
99.4
97.8
98.4
97.6
98.9
98.2
98.0
98.4
97.7
98.6
98.0
98.4
98.4
98.8
99.5
98.6
98.6
98.6
98.0
97.5
98.6
98.6
EXAMPLE 2: A sample of 100 observations is collected and yields x = 75 and  = 8.
Find a 95% confidence interval for the true population average.
EXAMPLE 3: Kennesaw State University claims the average starting salary of its
graduates is $38,500. A sample of 100 KSU students is sampled and yields
an average starting salary of $36,800 with a standard deviation of $9,369.
Using a 95% confidence level what can you say about KSU's claim?
EXERCISES:
In exercises 1 – 4, find the critical value z that corresponds to the given confidence level.
2
1. 98%
2. 95%
3. 96%
4. 99.5%
In exercises 5 – 8, use the given confidence interval and sample data to find (a) the
margin of error E and (b) a confidence interval for estimating the population mean .
5. Use a 99% confidence level with example 3.
6. Starting salaries of college graduates who have taken a statistics course: 95% confidence; n = 28,
x = $45,678, the population is normally distributed, and  is known to be $9900.
7. Ages of drivers occupying the passing lane whose speeds are less than the speed limit and with their
left signal flashing: 99% confidence interval; n = 50, x = 80.5 years, and  = 4.6 years
8. During TV commercials breaks, the time between uses of the remote control by males: 90% confidence; n = 25, x = 5.24 seconds, the population is normally distributed, and  = 2.50 seconds.
9. In order to monitor the ecological health of the Florida Everglades, various measurements are
recorded at different times. The bottom temperatures are recorded at the Garfield Bight station
and the mean of 30.4ºC is obtained for 61 temperatures on 61 different days. Assuming that
 = 1.7ºC, find a 95% confidence interval estimate of the population mean of all such
temperatures. What aspect of this problem is not realistic?
10. When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A
sample of 40 smokers has a mean cotinine level of 172.5. Assuming that  is known to be 119.5,
find a 90% confidence interval estimate of the mean cotinine level of all smokers.
11. The two intervals (114.4, 115.6) and (114.1, 115.9) are confidence intervals for  = true average
resonance frequency (in hertz) for all tennis rackets of a certain type. (a) What is the value of the
sample mean resonance frequency? (b) The confidence level for one of the intervals is 90% and for
the other it is 99%. Which is which and how can you tell?
12. An economist wants to estimate the first year’s mean income of college graduates who have had the
profound wisdom to take a statistics course. How many such incomes must be found if we want to be
95% confident that the sample mean is within $500 of the true population mean? Assume that a
previous study of such incomes has shown a  = $6250.
13. In order to keep from recalibrating their machinery, the U.S. Mint wants to
be 95% sure that their quarters have a mean of 5.65 grams. They collect the
weights of 30 quarters (to the right). If quarter’s weights standard deviation
is 0.068 grams, will they have to recalibrate?
5.63
5.68
5.62
5.60
5.53
5.58
5.60
5.58
5.59
5.66
14. SAT scores are normally distributed with a mean of 1000 and standard
deviation of 120. Mr. Markley’s calculus students earned the scores
shown to the right. Find the 99% confidence interval estimate of the
mean. What conclusion can be drawn concerning the SAT’s mean score
and the calculus class’s mean score?
940
1140
1040
1120
990
1070
1080
1190
5.73
5.59
5.63
5.66
5.67
5.60
5.74
5.57
5.62
5.73
5.60
5.60
5.57
5.71
5.62
5.72
5.57
5.70
5.60
5.49
850
1170
1330
1220
990
1120
1000
1100
Answers:
Body Temp Solution: First verify that the required assumptions are satisfied. The value of  is known (0.62ºF)
and the sample size, n = 50, is greater than 30. There are no outliers. (Because n > 30, there is no need to check
for normality; a histogram of the 50 temps, however, would show normality.) The required assumptions are
therefore satisfied and we can proceed.
n = 50, x = 98.28
a) The 0.95 confidence level implies that  = 0.05, so z = 1.96. Use the formula to calculate margin of error:

2
0.62
 0.171855
2
n
50
b) With x = 98.28 and E = 0.171855, we construct the confidence interval:
x –E<< x +E
98.28 – 0.171855 <  < 98.28 + 0.171855
98.11 <  < 98.45 (rounded to the nearest 100th as in x .
E  z
 1.96
INTERPRETATION: The result could be expressed as 98.28 ± 0.17 or (98.11, 98.45). Based on the sample
size of 50 and  = 0.62, the confidence interval for the population mean  is 98.11ºF <  < 98.45ºF and this
interval has a 0.95 confidence level. This means that if we were to select many different samples of size 50 and
construct the confidence intervals as we did here, 95% of them would actually contain the value of the
population mean . Note that our interval does not contain the value 98.6ºF, the value generally believed to be
the mean body temperature. Based on these results, it seems very unlikely that 98.6ºF is the correct value for
average body temperature.

8
 75  1.568 = (73.432, 76.568).
2
n
100

9369
EXAMPLE 3:  x  z
 36,800  1.96
 36,800  1,838.32  (34,963.68, 38,636.32)
2
n
100
Since KSU claimed that $38,500 is the mean starting salary & it falls in the interval, the claim
cannot be rejected.
EXAMPLE 2:  x  z
 75  1.96
1. 2.33
2. 1.96
3. 2.054
4. 2.807
5. a) 2,412.52, b) (34,387.48, 39,212.52)
6. a) 3,667.01, b) (42,010.99, 49,345.01)
7. a) 1.68, b) (78.82, 82.18)
8. a) 0.823, b) (4.42, 6.06)
9. (29.973, 30.827); it is unrealistic to know 
10. (141.4, 203.6)
11. a) 115.0, b) as the confidence level increases, the width of the interval increases, so 114.1 is in the 99%
12. 601
13. (5.602, 5.651), since 5.65, their desired mean, is included in the interval, they do not have to recalibrate.
14. (1007.1, 1161.6), since 1000 does not fall in the interval, you can be at least 99% sure that the calculus
students will score higher than the average score.