Unformatted text preview: STAT 2023  Holbrook Inferences About Means Chapter 23 23  1 STAT 2023  Holbrook Confidence Interval Estimate for the Mean (σ Known) 23  2 STAT 2023  Holbrook Recall: Confidence Limits for Population Mean Parameter =
Statistic ± Error Error = Y − µ or Y + µ (3) Y − µ Error
Z=
=
σy
σy (4) 23  3 µ = Y ± Error (2) © 19841994
T/Maker Co. (1) Error = Zσy (5) µ = Y ± Zσy STAT 2023  Holbrook Confidence Interval Mean (σ Known) 1. Assumptions 23  4 Population Standard Deviation Is Known
Population Is Normally Distributed
If Not Normal, Can Be Approximated by
If
Normal Distribution (n ≥ 30)
Normal Confidence Interval Mean (σ Known) STAT 2023  Holbrook 1. Assumptions Population Standard Deviation Is Known
Population Is Normally Distributed
If Not Normal, Can Be Approximated by
If
Normal Distribution (n ≥ 30)
Normal 2. Confidence Interval Estimate σ
σ
Y − Z⋅
≤µ ≤ Y + Z⋅
n
n 23  5 Important!
STAT 2023  Holbrook Rarely do we ever know what σ (the
Rarely
population standard deviation) is.
population 23  6 STAT 2023  Holbrook Example: Confidence Interval Estimate for the Mean (σ Known)
23  7 Thinking Challenge
STAT 2023  Holbrook You’re a Q/C inspector for
You’re
Gallo. The σ for 2liter bottles
is .05 liters. A random sample
.05
of 100 bottles showedY =
100
1.99 liters. What is the 90%
90%
confidence interval estimate of
the true mean amount in 2liter
mean
bottles?
2 liter
2 liter
© 19841994 T/Maker Co. 23  8 STAT 2023  Holbrook Confidence Interval Solution*
σ
σ
Y −Z⋅
≤µ≤ Y + Z⋅
n
n .05
.05
1.99 −1.645 ⋅
≤ µ ≤1.99 +1.645 ⋅
100
100
1.982 ≤ µ ≤ 1.998
23  9 Interpretation
STAT 2023  Holbrook I’m 90% confident that the population
I’m
mean (µ) amount of Gallo wine found in 2mean
liter bottles is between 1.982 and 1.998. 23  10
10 STAT 2023  Holbrook Confidence Interval Estimate for the Mean (σ Unknown) 23  11
11 STAT 2023  Holbrook Confidence Interval Mean (σ Unknown) 1. Assumptions Population Standard Deviation Is Unknown
Population is Normally Distributed
Population Normally
If Not Normal, Can Be Approximated by Normal
If
Distribution (n ≥ 30)
Distribution 2. Use t Distribution 23  12
12 Confidence Interval Mean (σ Unknown) STAT 2023  Holbrook 1. Assumptions
Population Standard Deviation Is Unknown
Population is Normally Distributed
Population Normally
If Not Normal, Can Be Approximated by Normal
If
Distribution (n ≥ 30)
Distribution 2. Use t Distribution
3. Confidence Interval Estimate S
S
Y − t n −1 ⋅
≤ µ ≤ Y + t n −1 ⋅
n
n 23  13
13 Checking Assumptions
STAT 2023  Holbrook 1. When the book asks you to check the assumptions this
When
is what you need to do:
is Check and see if the population of whatever you are
Check
measuring is Normally Distributed.
measuring If you don’t know or can’t determine if the population is
If
normally distributed, check the sample size.
normally 23  14
14 If it is normal (or if the histogram of your sample data is “mound
If
shaped”) then you can conduct the test.
can If n ≥ 3 0 then central limit theorem applies and we can conduct
If
can
the test.
the
If n < 30 and the population is not normal, then we can not conduct
If
can
the test.
the STAT 2023  Holbrook An Aside:
Practice Using the T Table
(Appendix D) 23  15
15 Find the T Value
STAT 2023  Holbrook For a 90% C.I. and a sample of size n=3,
For
find the t value.
find 23  16
16 STAT 2023  Holbrook Degrees of Freedom (df)
df = n1 1. Number of Observations that Are Free to
Number
Vary After Sample Statistic Has Been
Calculated
Calculated
2. Example 23  17
17 Sum of 3 Numbers Is 6
X1 = 1 (or Any Number)
(or
X2 = 2 (or Any Number)
(or
X3 = 3 (Cannot Vary)
(Cannot
Sum = 6 degrees of freedom
degrees
= n 1
= 3 1
1
=2 T Table
STAT 2023  Holbrook 90% C.I. 95% C.I. df Two tail probability
(what we use for C.I.’s) .20 .10 .05 .10 .05 .025 1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182 t values
23  18
18 One tail probability
One Assume:
n=3
df = n  1 = 2
90% C.I.
So, α = .10
So, T Table
STAT 2023  Holbrook 90% C.I. 95% C.I.
90% df Two tail probability
(what we use for C.I.’s) .20 .10 .05 .10 .05 .025 1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182 23  19
19 Assume:
n=3
df = n  1 = 2
90% C.I.
So, α = .10
So, T Table
STAT 2023  Holbrook 90% C.I. 95% C.I.
90% df Two tail probability
(what we use for C.I.’s) .20 .10 .05 .10 .05 .025 1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182 23  20
20 Assume:
n=3
df = n  1 = 2
90% C.I.
So, α = .10
So,
.10 T Table
STAT 2023  Holbrook 90% C.I. 95% C.I.
90% df Two tail probability
(what we use for C.I.’s) .20 .10 .05 .10 .05 .025 1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182 23  21
21 Assume:
n=3
df = n  1 = 2
90% C.I.
So, α = .10
So, Find the T Value Solution
STAT 2023  Holbrook So the t value for a 90% C.I. and a sample
So
of size n=3 is t = 2.92.
2.92 23  22
22 Note about the T Table
STAT 2023  Holbrook Note that the infinity line of the T Table
Note
represents the Zvalues we used for
confidence intervals in Chapter 19 and
Chapter 22.
Chapter 23  23
23 STAT 2023  Holbrook Example: Confidence Interval Estimate for the Mean (σ unknown)
23  24
24 C.I. For Mean (σ unknown) STAT 2023  Holbrook How well do students perform
How
academically? 101 students at
101
SFCC were asked what their
GPA was. The sample mean
was 3.13 and the sample
3.13
standard deviation was
0.4767. Calculate and
interpret a 95% confidence
95%
interval estimate for population
mean.
mean.
23  25
25 Assumptions
STAT 2023  Holbrook Note: We don’t know if the population is
normal, but n ≥ 30 so the central limit
theorem applies and we can conduct the
test.
test. 23  26
26 Find the T Value
STAT 2023  Holbrook For a 95% C.I. and a sample of size
For
n=101, find the t value.
n=101, 23  27
27 Find the T Value Solution
STAT 2023  Holbrook The t value for a 95% C.I. and a sample of
The
size n=101, or df=100 is t = 1.984.
1.984 23  28
28 STAT 2023  Holbrook Confidence Interval Solution*
s
s
Y−t⋅
≤µ≤Y+t⋅
n
n 0.4767
0.4767
3.13 − 1.984 ⋅
≤ µ ≤ 3.13 + 1.984 ⋅
101
101
3.036 ≤ µ ≤ 3.224
23  29
29 Interpretation
STAT 2023  Holbrook I’m 95% confident that the population
I’m
mean (µ) GPA of all SFCC students is
mean
GPA
between 3.036 and 3.224.
between 23  30
30 Confidence Interval Solution STAT 2023  Holbrook 1. Using the TI83 Hit “STAT” scroll to “TESTS” then “8”, hit
”,
“ENTER” “TInterval“ Select “Stats” hit “ENTER” then enter the
then
appropriate statistics
appropriate For example: y = 3.13, S=0.4767, n=101 23  31
31 Choose a “.95” “CLevel”
Select “Calculate” and hit “ENTER” STAT 2023  Holbrook Hypothesis Test for the Mean (σ Known) 23  32
32 Important!
STAT 2023  Holbrook Rarely do we ever know what σ (the
Rarely
population standard deviation) is.
population 23  33
33 STAT 2023  Holbrook Hypothesis Test for Mean (σ Known) 1. Assumptions 23  34
34 Population Standard Deviation Is Known
Population Is Normally Distributed
If Not Normal, Can Be Approximated by
If
Normal Distribution (n ≥ 30)
Normal STAT 2023  Holbrook Hypothesis Test for Mean (σ Known) 1. Assumptions Population Standard Deviation Is Known
Population Is Normally Distributed
If Not Normal, Can Be Approximated by
If
Normal Distribution (n ≥ 30)
Normal 2. ZTest Statistic
Y − µy
Y −µ
Z=
=
σ
σy
n
23  35
35 Note:
STAT 2023  Holbrook We will not work examples for this type of
We
not
hypothesis test because it is so rare that
we know what σ is (and that none of your
homework problems give you !!)
!!) 23  36
36 STAT 2023  Holbrook Hypothesis Test of the Mean (σ Unknown) 23  37
37 STAT 2023  Holbrook Hypothesis Test for Mean (σ Unknown) 1. Assumptions 23  38
38 Population Standard Deviation Is Unknown
Population is Normally Distributed
Population Normally
If Not Normal, Can Be Approximated by Normal
If
Distribution (n ≥ 30)
Distribution STAT 2023  Holbrook Hypothesis Test for Mean (σ Unknown) 1. Assumptions Population Standard Deviation Is Unknown
Population is Normally Distributed
Population Normally
If Not Normal, Can Be Approximated by Normal
If
Distribution (n ≥ 30)
Distribution 2. Use t Distribution 23  39
39 STAT 2023  Holbrook Hypothesis Test for Mean (σ Unknown) 1. Assumptions Population Standard Deviation Is Unknown
Population is Normally Distributed
Population Normally
If Not Normal, Can Be Approximated by Normal
If
Distribution (n ≥ 30)
Distribution 2. Use t Distribution
3. t Test statistic 23  40
40 Y −µ
t=
s
n Checking Assumptions (Same steps as for Confidence Intervals) STAT 2023  Holbrook 1. When the book asks you to check the assumptions this
When
is what you need to do:
is Check and see if the population of whatever you are
Check
measuring is Normally Distributed.
measuring If you don’t know or can’t determine if the population is
If
normally distributed, check the sample size.
normally 23  41
41 If it is normal (or if the histogram of your sample data is “mound
If
shaped”) then you can conduct the test.
can If n ≥ 3 0 then central limit theorem applies and we can conduct
If
can
the test.
the
If n < 30 and the population is not normal, then we can not conduct
If
can
the test.
the STAT 2023  Holbrook Note: Step 3 – finding the pvalue 1. We must use our TI83 to find pvalues (the Z
We
table will not work as it did in Chapter 20).
table 23  42
42 STAT 2023  Holbrook Example:
Hypothesis Test for the Mean (σ Unknown) 23  43
43 STAT 2023  Holbrook Hypothesis Test for the Mean Example Is the average capacity of
Is
batteries at least 140 ampereamperehours? A random sample of 20
hours?
20
batteries had a mean of 138.47
138.47
& a standard deviation of 2.66.
2.66
Assume that the population of
battery capacities are normally
distributed.
distributed. 23  44
44 STAT 2023  Holbrook H0: µ = 140
Ha: µ < 140
n = 20
df = 20 – 1 = 19
Test Statistic: t Test Solution
Pvalue:
Pvalue: Decision:
Decision:
Conclusion: 23  45
45 STAT 2023  Holbrook t Test Solution Step 2: Test Statistic Y − µ 138.47 − 140
t=
=
= −2.57
S
2.66
n
20 23  46
46 STAT 2023  Holbrook t Test Solution H0: µ = 140
Ha: µ < 140
n = 20
df = 20 – 1 = 19
Test Statistic: t = −2.57 Pvalue: Decision:
Conclusion: 23  47
47 t Test Solution : Finding the pvalue STAT 2023  Holbrook 1. Using the TI83 Hit “2nd” and then “VARS” (Bringing up “DISTR”)
Scroll down to 5, “tcdf(“ and hit “ENTER”
Scroll “tcdf(“ should appear
Hit “” then “1” then “2nd” then “,” then “99” 23  48
48 The default is tcdf(lower t value, upper t value, df) This is the Lower bound = 1 x 1099 (negative infinity) Separate lower bound by a “,”
Enter “2.57” separated by a “,”
Enter “19” for df and then “)”
Screen should read tcdf(1E99, 2.57, 19)
Screen
tcdf(1 STAT 2023  Holbrook t Test Solution H0: µ = 140
Ha: µ < 140
n = 20
df = 20 – 1 = 19
Test Statistic: t = −2.57 Pvalue: =0.0094
Pvalue: =0.0094
0.94% of the time we
0.94%
would have seen
data like this if H0
data
was true. (rarely)
was
Decision:
Conclusion: 23  49
49 STAT 2023  Holbrook t Test Solution H0: µ = 140
Ha: µ < 140
n = 20
df = 20 – 1 = 19
Test Statistic: t = −2.57 Pvalue: =0.0094
Pvalue: =0.0094
0.94% of the time we
0.94%
would have seen
data like this if H0
data
was true. (rarely)
was
Decision:
Reject H0
Conclusion: 23  50
50 STAT 2023  Holbrook t Test Solution H0: µ = 140
Ha: µ < 140
n = 20
df = 20 – 1 = 19
Test Statistic: t = −2.57 Pvalue: =0.0094
Pvalue: =0.0094
0.94% of the time we
0.94%
would have seen
data like this if H0
data
was true. (rarely)
was
Decision:
Reject H0
Conclusion: 23  51
51 Enough evidence that the
Enough
population mean battery capacity
is less than 140 ampere hours.
is STAT 2023  Holbrook Example:
Hypothesis Test for the Mean (σ Unknown) 23  52
52 STAT 2023  Holbrook TwoTailed t Test Example Does an average box of
Does
cereal contain 368
368
grams of cereal? A
random sample of 36
36
boxes had a mean of
372.5 & a standard
372.5
deviation of 12 grams.
deviation 12 grams.
368 gm.
23  53
53 STAT 2023  Holbrook H0: µ = 368
Ha: µ ≠ 368
n = 36
df = 36 – 1 = 35
Test Statistic: t Test Solution
Pvalue:
Pvalue: Decision:
Decision:
Conclusion: 23  54
54 STAT 2023  Holbrook t Test Solution Step 2: Test Statistic Y − µ 372.5 − 368
t=
=
= 2.25
S
12
n
36 23  55
55 STAT 2023  Holbrook H0: µ = 368
Ha: µ ≠ 368
n = 36
df = 36 – 1 = 35
Test Statistic: t = 2.25 t Test Solution
Pvalue:
Pvalue: Decision:
Decision:
Conclusion: 23  56
56 STAT 2023  Holbrook TwoTailed t Test pValue Solution pvalue is P(t ≤ 2.25 or t ≥ 2.25)
pvalue
1/2 pValue 1/2 pValue 2.25
1.50 0 2.25
1.50 23  57
57 Zt
t value of sample
value
statistic (observed)
statistic t Test Solution : Finding the pvalue STAT 2023  Holbrook 1. Using the TI83 (LOWER TAIL ONLY!) Hit “2nd” then “VARS” then “5”
“tcdf(“ should appear
Hit “” then “1” then “2nd” then “,” then “99” 23  58
58 This is the Lower bound = 1 x 1099 (negative infinity) Separate lower bound by a “,”
Enter “2.25” separated by a “,”
Enter “35” for df and then “)”
Screen should read tcdf(1E99, 2.25, 35)
Screen
tcdf(1 STAT 2023  Holbrook TwoTailed t Test pValue Solution pvalue is P(t ≤ 2.25 or t ≥ 2.25)
pvalue
.0154
1/2 pValue 1/2 pValue 2.25
1.50 0 2.25
1.50 23  59
59 Zt
t value of sample
value
statistic (observed)
statistic STAT 2023  Holbrook TwoTailed t Test pValue Solution
Due to symmetry
. 0154
1/2 pValue 1/2 pValue
. 0154  2.25
1.50 0 2.25
1.50 23  60
60 Zt
t value of sample
value
statistic (observed)
statistic STAT 2023  Holbrook TwoTailed t Test pValue Solution So the pvalue = .0154 + .0154 = .0308
So
.0154
1/2 pValue 1/2 pValue
.0154  2.25
1.50 0 2.25
1.50 23  61
61 Zt
t value of sample
statistic (observed)
statistic STAT 2023  Holbrook H0: µ = 368
Ha: µ ≠ 368
n = 36
df = 36 – 1 = 35
Test Statistic: t = 2.25 t Test Solution
Pvalue: =0.0308
Pvalue: =0.0308
3.08% of the time we
3.08%
would have seen data
like this if H0 was true.
like
(rarely)
(rarely)
Decision:
Conclusion: 23  62
62 STAT 2023  Holbrook H0: µ = 368
Ha: µ ≠ 368
n = 36
df = 36 – 1 = 35
Test Statistic: t = 2.25 t Test Solution
Pvalue: =0.0308
Pvalue: =0.0308
3.08% of the time we
3.08%
would have seen data
like this if H0 was true.
like
(rarely)
(rarely)
Decision:
Reject H0
Conclusion: 23  63
63 STAT 2023  Holbrook H0: µ = 368
Ha: µ ≠ 368
n = 36
df = 36 – 1 = 35
Test Statistic: t = 2.25 t Test Solution
Pvalue: =0.0308
Pvalue: =0.0308
3.08% of the time we
3.08%
would have seen data
like this if H0 was true.
like
(rarely)
(rarely)
Decision:
Reject H0
Conclusion: 23  64
64 Enough evidence that the
Enough
population mean amount of cereal
found in Cereal O’s is different
from 368 grams. Hypothesis Test Solution
Alternate Method STAT 2023  Holbrook 1. Using the TI83 Hit “STAT” scroll to “TESTS” then “2”, hit “ENTER” “TTest“ Select “Stats” then hit “ENTER“
Scroll to μ0 and enter the hypothesized value
Scroll
the
For example: μ0=368
For Enter the appropriate statistics For example: y = 372.5, S=12, n=36 Scroll and select (by hitting “ENTER”) one of the
”)
following: two tail (≠ ), lower tail (<) or upper tail (>)
following:
),
alternative hypothesis.
alternative
For example: ≠ μ0
For Select “Calculate” and hit “ENTER” 23  65
65 STAT 2023  Holbrook Finding Sample Sizes
(Note: we skipped this topic with respect to proportions.) 23  66
66 STAT 2023  Holbrook (1)
(2)
(3) Finding Sample Sizes for Estimating µ Y − µ Error
Z=
=
σy
σy
σ
s
Error = Zσ y = Z
≅Z
n
n
22
Zs
n=
2
(Error )
Error Is Also Called Bound, B
Error
23  67
67 I don’t want to
sample too much
or too little! Note
STAT 2023  Holbrook • Note: This formula and process of
finding n (the samples size) is slightly
different (and easier) and more
conservative than the formula and
procedure described in your book on
Page 603. (Page 442 – First Edition.)
(Page 23  68
68 Sample Size Example
STAT 2023  Holbrook What sample size is needed to be 90%
What
confident of being correct within a bound
or error of ± 5? A pilot study suggested
that the standard deviation is 45.
that 23  69
69 Sample Size Example
Solution STAT 2023  Holbrook (1.645) ( 45) = 219.2 ≅ 220
Zs
n=
=
2
2
Error
( 5)
22 2 2 Always round up!
(Yes, I know it goes against everything that you’ve learned!) 23  70
70 End of Chapter
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 Statistics, Normal Distribution, Holbrook

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