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Transcript
With 𝜎 unknown


Problem 1: Given a sample mean of 2.8 with a
sample size of 34, and standard deviation of
.05; find the 95% confidence interval for the
mean of the population.
Problem 2: Given a sample proportion of 28%
with a sample size of 34, find the 95%
confidence interval for the proportion of the
population.

We took data and used the sample standard
deviation along with the process for finding a
confidence interval with 𝜎 known …
β—¦ However, we don’t really know 𝝈 in that case (we
know s), so this is actually not the best way to
construct a confidence interval in that situation!
β—¦ Instead, we can use a different distribution along
with s in order to be more precise…

Created for use in small samples and with
standard deviation unknown

Considers β€œdegrees of freedom”
β—¦ If you have a average of 100 for a population of 10,
the first 9 items are free to be whatever they want,
but the 10th item must make the sum be 1,000, so
it is not free. Therefore a population of 10 has 9
degrees of freedom

We will always choose n – 1 degrees of
freedom

You will find the value of π’•πœΆ/𝟐 in table A3 as
part of the process of calculating E in the
confidence interval for the mean. For this
you need:
β—¦ Degrees of freedom: n – 1
β—¦ Sum of the areas of the 2 tails (which is equal to 𝛼)
ο‚– Remember, 𝛼 = 1 βˆ’ confidence level desired

Find 𝑑𝛼 for a confidence level of 95% and a
2
sample size of 35

Find 𝑑𝛼 for a confidence level of 90% and a
2
sample size of 40

Find 𝑑𝛼 for a confidence level of 80% and a
2
sample size of 1001



Unless the original distribution is already
normal, we must have a sample size of
n > 30.
If normal, there is no size requirement
This method is only useful when 𝜎 is not
known for the population


Use z
n > 30 and 𝜎 is known
β—¦ Data is normally distributed and 𝜎 is known


Use t
n > 30 and 𝜎 is not known
β—¦ Data is normally distributed and 𝜎 is not known

If n is 30 or less and the data is not normally
distributed, neither method applies.

Just as it was in 7.3, the formula for the
confidence interval is the sample statistic ± E,
however, the way we calculate E changes.

Confidence Interval: (π‘₯ βˆ’ 𝐸, π‘₯ + 𝐸)

Formula for E:
𝐸=𝑑 βˆ™
𝛼
2
𝑠
𝑛


The average incomes of the families of 36
randomly selected Bradley University
students are listed below:
28
29
35
42
42
68
44
50
52
54
78
74
84
90
95
101
108
88
116
121
122
158
167
120
235
59
150
56
133
51
75
79
47
88
180
26
Determine a 95% confidence interval for the
actual average income of a Bradley family
P.365-367:
#5-12 (simply state whether you would use t,
z, or neither)
 #19-20


5. t
6.z
7. Neither
8.t
9. z
10. Neither
11. t
12.t
19. a.) 98.20ºF
b.) 98.04ºF < m < 98.36ºF because the
confidence interval limits do not contain 98.6ºF, the
results suggest that the mean is less than 98.6ºF.
20. a.) 2.1 lb
b.) 0.0 lb < m < 4.2 lb
c.) There does appear to be a loss of weight,
but the actual amount of lost weight is so small that
the weight loss program does not appear to be
practical.

46 subjects were treated with raw doses of
garlic to see if it lowered their cholesterol.
The change in was measured and resulted in
the following statistics:
β—¦ 𝑛 = 46, π‘₯ = 0.4, 𝑠 = 21.0

Create a 95% confidence interval to estimate
the actual average drop in cholesterol from
doses of garlic.

Complete and finish the worksheet.

****SKIP 2 and 5!


What you do not finish is for HOMEWORK!
Yes you need to show your work!
Go back to the project you have been working
on. Since you now know your confidence
intervals were slightly off, update them using the
new methods we learned today (do not erase any
of the work you’ve done so far, and keep the
answers you gave to the questions).


Answer the following additional question in a 4-5
sentence paragraph:
β—¦ What changes did you notice in your results after now
completing the process in the correct way? Was the
change enough for you to feel it was worth redoing the
calculations?