9.2 Examples
Example 1
A simple random sample of size n is drawn. The sample mean, , is found to be 19.2, and the sample
standard deviation, s, is found to be 4.7.
a) Construct a 95% confidence interval about µ (population mean) if the sample size, n, is 35.
READ pages 405 – 412
However, we will use our calculator!! ☺
On your calculator,
Go to STAT, then TESTS, then scroll down to TInterval.
You will have a choice of “data” or “stats”. We will choose stats since we know the mean and
standard deviation, rather than having a list of data values.
gives you this:
So the interval, rounded to two decimal places, is (17.59, 20.82)
b) Construct a 95% confidence interval about µ (population mean) if the sample size, n, is 51.
gives you this:
So the interval, rounded to two decimal places, is (17.88, 20.52)
How does increasing the sample size affect the margin of error, E?
Since you are DIVIDING by the square root of n, as n (the sample size) gets larger, it will cause
this value to decrease. (Dividing by a larger number gives you a smaller answer!)
So, the margin of error decreases as n gets larger.
c) Construct a 99% confidence interval about µ (population mean) if the sample size, n, is 35.
gives you this:
So the interval, rounded to two decimal places, is (17.03, 21.37)
d) If the sample size is 16, what conditions must be satisfied to compute the confidence interval?
Example 2
A trade magazine routinely checks the drive-through service times of fast-food restaurants. A
90% confidence interval that results from examining 519 customers in one fast food chain’s
drive-through has a lower bound of 177.0 seconds and an upper bound of 180.0 seconds. What
does this mean?
One can be 90% confident that the mean drive-through service time of this chain is between
177.0 and 180.0 seconds.
Example 3
In a survey conducted by a polling company, 1100 adult Americans were asked how many hours they
worked in the previous week. Based on the results, a 95% confidence interval for the mean numbers of
hours worked had a lower boung of 42.7 and an upper bound of 44.5. Provide two recommendations
for increasing the precision of the interval.
**Decrease the confidence level.
**Increase the sample size.
(Read pages 407-408)
Example 4
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose
for a random sample of 960 people age 15 or older, the mean amount of time spent eating or drinking
per day is 1.62 hours with a standard deviation of 0.65 hour.
a) A histogram of time spent eating/drinking each day is skewed right. Use this result to explain
why a large sample is needed to construct a confidence interval for the mean.
*Since the distribution is skewed right (not normally distributed) the sample must be large so
that the distribution of the same mean will be approximately normal.
b) In 2010, there were over 200 million people nationally age 15 or older. Explain why this, along
with the fact that our data is from a random sample, satisfies the requirements for constructing
a confidence interval.
*The sample size is less than 5% of the population (n < 0.05N)
c) Determine and interpret a 95% confidence interval for the mean amount of time Americans age
15 or older spend eating/drinking each day.
gives us this:
Rounded to two places, the interval is (1.58, 1.66)
Which means the nutritionist can be 95% confident that the mean amount of time Americans spend
eating/drinking per day is between 1.58 and 1.66 hours.
d) Could this interval be used to estimate the mean amount of time a 9-year-old spends eating and
drinking per day?
*No, the interval is about people age 15 and older. The amount for 9-year-olds could easily be
very different.
Example 5
A poll conducted in 1970 asked 1007 people, “During the past year, about how many books, either
hardcover or paperback, did you read either all or part of the way through?” Results of this survey
indicated that = 18.9 books and s = 19.9 books.
a) Construct a 95% confidence interval for the mean number of books read all or part of the way
through during the preceding year and interpret it.
gives you this:
Rounded to two places, the interval is (17.67, 20.13)
So we are 95% confident that the mean number of books read is between 17.67 and 20.13.
b) Compare these results to a recent survey of 1007 people. The results indicated that = 12.7
books and s = 15.9 books. A 95% confidence interval for this survey is (11.72, 13.68). Were
people reading more in 1970 than they are today?
*Yes. We are 95% confident that the mean number of books read is between 11.72 and 13.68
instead of 17.67 and 20.13. This is a decrease.
Example 6
The trade volume of a stock is the number of shares traded on a given day. The following data, in
millions (so that 2.45 represents 2,450,000 shares), represent the volume of a certain stock traded for a
random sample of 40 trading days in 2007.
a) Use this data to compute a point estimate for the population mean number of shares traded per
day in 2007.
Enter this data into L1 in your calculator.
and so on . . .
1-var stats gives you
= 2.338 (rounded to three places)
And also note that s = 1.215
b) Construct a 90% confidence interval for the population mean number of shares traded per day
in 2007.
gives you this:
So the interval rounded to three places is (2.014, 2.662)
So there is a 90% confidence that the mean number of shares of the stock traded per day in 2007 is
btween 2,014,000 and 2,662,000.
c) A second random sample of 40 days in 2007 resulted in the following data. Construct a 90%
confidence interval for the population mean number of shares traded per day in 2007.
Follow the same steps as in part b above. Enter this data into L1, find the 1-var stats to determine
and s so that you can do the TInterval on your calculator.
This interval you should obtain is (2.248, 3.111).
The two confidence intervals are different because of variation in sampling. The samples have different
means and standard deviations that lead to different confidence intervals.
Example 7
People were polled on how many books they read the previous year. Initial survey results indicate that
s = 11.2 books.
a) How many subjects are needed to estimate the number of books read the previous year within
six books with 90% confidence?
E = 6 (“within 6 books”)
The critical value z
α/2
that corresponds to 90% confidence is 1.645 (see page 395)
n = [ (1.645*11.2)/6]
2
= 3.07066666
2
= 9.42899 (we always round up because we are talking about a minimum sample size)
Therefore the 90% confidence level requires 10 subjects.
b) How many subjects are needed to estimate the number of books read the previous year within
three books with 90% confidence?
E = 3 (“within 3 books”)
The critical value z
α/2
that corresponds to 90% confidence is 1.645 (see page 395)
n = [ (1.645*11.2)/3]
2
= 6.1413333
2
= 37.715975 (we always round up because we are talking about a minimum sample size)
Therefore the 90% confidence level requires 38 subjects.
c) What effect does doubling the required accuracy have on the sample size?
Doubling the required accuracy quadruples the required sample size. (A sample size 38 is
almost 4 times larger than a sample size of 10. Increasing the accuracy needed will always
increase the sample size needed. )
d) How many subjects are needed to estimate the number of books read the previous year within
six books with 99% confidence?
E = 6 (“within 6 books”)
The critical value z
α/2
that corresponds to 99% confidence is 2.575 (see page 395)
n = [ (2.575*11.2)/6]
2
= 4.806666
2
= 23.1040444 (we always round up because we are talking about a minimum sample size)
Therefore the 99% confidence level requires 24 subjects.
Compare this result to part (a). How does the increasing level of confidence in the estimate affect
sample size? Why is this reasonable?
Increasing the level of confidence increases the sample size required. For a fixed margin of error,
greater confidence can be achieved with a larger sample size.
Example 8
Enter the data into L1, L2, L3 on your calculator.
̅ = 99.125 Sx = 18.4114677 n = 8
̅ = 99.15 Sx = 15.54035494 n = 20
̅ = 99.2 Sx = 14.73770674 n = 30
b)
answer: (83.73, 114.52)
answer: (91.88, 106.42)
answer: (93.70, 104.7)
