程序代写代做代考 matlab Hive ESS116

ESS116

ESS 116
Introduction to Data Analysis in Earth Science

Image Credit: NASA
Instructor: Mathieu Morlighem
E-mail: mmorligh@uci.edu (include ESS116 in subject line)
Office Hours: 3218 Croul Hall, Friday 2:00 pm – 3:00 pm
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Part 1: 60 min “in class” next week
May 12th, 2:00 pm
cheat sheet (Letter paper, double sided, hand written)
mix of multiple choice and short answer questions
Everything from Lecture 1 to 5 (no hypothesis testing)
Bring a scientific calculator
Part 2: 2 hours “in lab”
May 13th during your regular lab session
Open Book
Similar to what you have experienced with your labs
No make up exam for the midterm
Midterm exam

Lecture 5 quick review

Lecture 6 – Hypothesis testing
Sampling Distribution of the sample mean
Central Limit Theorem (CLT)
Confidence intervals
Hypothesis Testing
t-test (Comparing means)
χ2-test (Goodness of fit)

Today’s lecture

Population: the actual properties of the real world
Sample: set of values imperfectly representing the population

Parameters: refer to the population (e.g., μ and σ)
Statistics: refer to the sample (e.g., and s)

Accuracy: quality of being close to the true value
Precision: number of significant digits in a numerical value (measurements or calculation)
Lecture 5 – review

Sample visualization
Frequency Table
Cumulative Frequency
Histogram
Rules for a good histogram
number of bins ≈ sample size
histogram takes either a number
of bins, or a list of bin edges

Lecture 5 – review

Central Tendency:
Mean (average)
Median (50% higher, 50% lower)
Mode(s) (peak value(s))
Dispersion:
Range (max – min)
Standard deviation (average distance to mean)
Shape:
Skewness (positive: tail to the right, negative: tail to the left)
Kurtosis (<3: flat, >3: peaked)
Know how they relate to visual features on a histogram

Statistical parameters
what you need to remember

Histograms: empirical frequency distribution of our sample.
A histogram for and an infinitely small bin size will produce a Probability Density function (PDF)
The probability that x is between x1 and x2 is:

Examples of theoretical Distributions:
Normal distribution (2 parameters: μ and σ)
Z distribution (0 parameters)
Student’s t distribution (1 parameter: Φ)

Probability Density Functions

Normal (μ,σ)
Given x0, find p0
>> p0 = normcdf(x0,mu,sigma);
Given p0, find x0
>> x0 = norminv(p0,mu,sigma);
Z-distribution
>> p0 = normcdf(x0);
>> x0 = norminv(p0);
t-distribution
>> p0 = tcdf(x0,V);
>> x0 = tinv(p0,V);
χ2-distribution
>> p0 = chi2cdf(x0,V);
>> x0 = chi2inv(p0,V);

MATLAB theoretical distributions

p0 = P( x < x0) e.g.: 0.88 = P(x < 1.17) ESS 116 grades follow a Normal distribution of mean 800 with a standard deviation of 100. What is the probability of having a grade below 500? 1 – normcdf(500,800,100); 1 – norminv(500,800,100); normcdf(500,800,100); norminv(500,800,100); i>Clicker question

Lecture 6 – Hypothesis testing

Central Limit Theorem

Central Limit Theorem

Central Limit Theorem

Population distribution

Sampling distribution of the sample mean
Sample size

Central Limit Theorem in Action

The average male drinks 2 L of water when active outdoors (with a standard deviation of 0.7 L). You are planning a full day in nature with 50 men and will bring 110 L of water. What is the probability that you run out ?
Example

Example
Sampling distribution of the Sample mean
P(run out) = P(average water use > 110/5 L)

= P(average water use > 2.2 L)

= P( > 2.2)

= 1 – P( < 2.2) = 1 – normcdf(2.2,2,0.7/sqrt(50)) = 0.0217 The probability of running out of water is 2.17% Population distribution Confidence Intervals Confidence Intervals: provide statistical limits for your mean values based on a degree of statistical confidence. Ex: We can say with 95% confidence that the average temperature in Irvine is within [18°C 24°C] or 21 ±3°C How to calculate this interval? Set the level of significance α (α = 0.05 for a 95% Confidence Interval) Find such that Confidence interval in the mean Using the Central Limit Theorem 1- α (95%) α/2 α/2 Find such that: Normal Distribution Normalizing (Subtract divide by ) 1- α (95%) α/2 α/2 Find such that: 1 0 Z-Distribution What if we don’t know σ, can we say σ ≈ s ? If the sample size n > 30: Yes
If the sample size n < 30: Yes but the distribution of X needs to be roughly normal we pay a penalty: use a t-distribution instead of a z-distribution (fatter tails) Distribution of the sample mean Set the level of significance α (α = 0.05 for a 95% CI) follows a: z-distribution if n>30
t-distribution if n<30 (Φ = n-1) Find such that: Confidence interval in the mean Sampling distribution of the Sample mean – “Normalized” 0 - 1- α α/2 α/2 if n<30: deltax = -tinv(alpha/2,n-1)*s/sqrt(n); Or (equivalent) deltax = tinv(1-alpha/2,n-1)*s/sqrt(n); Else: deltax = -norminv(alpha/2)*s/sqrt(n); Or (equivalent) deltax = norminv(1-alpha/2)*s/sqrt(n); You sample 36 apples from your farm’s harvest of over 200,000 apples. The mean weight of the sample is 112 grams (with s = 40 grams). What is the probability that the mean weight of the 200,000 apples is within 100 and 124 grams? (i.e. mean is 112±12 g) Example n>30

P(μ within 12 of ) = P( within 12 of μ)

= P( within 12 of )

= P( )

= normcdf(12/(40/6))
– normcdf(-12/(40/6))

= 0.9281
Example n>30

We have a 92.8% chance that the actual mean is within 12 grams of our sample mean

Sampling distribution of
the Sample mean – “Normalized”

0

7 patients’ blood pressures have been measured after having been given a new drug for 3 months. They had blood pressure increases of 1.5, 2.9, 0.9, 3.9, 3.2, 2.1 and 1.9

Construct the 95% Confidence Interval (CI) for the true expected blood pressure increase for all patients in a population.
Example n<30 Example n<30 Sampling distribution of the Sample mean (Student’s t distribution with Φ = n-1 = 6) Here are our statistics for n = 7 What is our 95% confidence interval ? 95% 2.5% 2.5% 0 -Δz = tinv(0.025,7-1) = -2.4469 Δz = tinv(0.975,7-1) = +2.4469 - Δz Δz There is a 95% chance that the mean, μ, is within 2.3429 ± 0.9639 %Compute sample size, sample mean and standard deviation n    = length(data); xbar = mean(data); s    = std(data);   %Set level of significance (95% CI) alpha = 0.05;   %Depending on the sample size, we either use a z or t distrib. if n<30     deltax = tinv(1-alpha/2,n-1)*s/sqrt(n); else     deltax = norminv(1-alpha/2)*s/sqrt(n); end fprintf('Population mean is %g ± %g (95%% CI)\n',xbar,deltax); MATLAB code for confidence Interval Hypothesis Testing You read that, on average, a volcanic eruption lasts 7 weeks (μ=7). But we suspect that this number is wrong and should higher (μ>7).

How can we prove this, for a given level of significance (α=0.05)?

We look at the past n=100 eruptions and find =7.2 and s =1 week.
Introduction

Assuming that μ = 7, there is only a 2.3% chance of finding a mean of 7.2 weeks, so we can reject μ=7
Conclusion: μ>7

P( ≥ 7.2) = 1 – P( < 7.2) = 1 – normcdf(7.2,7,1/10) = 0.0228 < α Assuming that μ=7 P( ≥ 7.2) = 1 – P( < 7.2) = 1 – normcdf(7.2,7,1/10) = 0.0228 < α You read that, on average, a volcanic eruption lasts 7 weeks (μ=7). But we suspect that this number is wrong and should higher (μ>7).

How can we prove this, for a given level of significance (α=0.05)?

We look at the past n=100 eruptions and find =7.2 and s =1 week.
Testing one population mean

Null Hypothesis H0
Alternative Hypothesis H1
p-value
Assuming that μ = 7, there is only a 2.3% chance of finding a mean of 7.2 weeks, so we can reject μ=7
Conclusion: μ>7

Assuming that μ=7

Null and alternative hypotheses:
Prepare a statement about a fact for which it is possible to calculate its probability of occurrence (e.g.: μ=7, μ1> μ2, etc.)
This statement is the null hypothesis, H0, and its counterpart is the alternative hypothesis, H1.
H0 is often the reverse of what the experimenter actually believes for tactical reasons.

Level of significance (α): If P(H0) < α, we reject H0 p-value: the probability of rejecting H0, while H0 is actually true If p-value > α, H0 is not rejected
The lower the p-value, the stronger is the evidence provided by the data against the null hypothesis.

Hypothesis testing

t-test “Comparing Means”

Unpaired t-test
the two samples are from independent populations
Ex1: Are tropical fish larger than temperate fish?
Ex2: Are the temperatures in Long Beach and Death Valley significantly different?

Paired vs Unpaired t-test
Paired t-test
the two samples are from the same population
Ex1: Do fish get larger as they age?
Ex2: Is the annual temperature in the last 5 years in Death Valley significantly higher than in the Earlier 5 years?

Unpaired t-test (independent populations)
Sample 1: size n1, mean m1 and standard deviation s1
Sample 2: size n2, mean m2 and standard deviation s2

Paired t-test (same population)
Both samples should have the same size, n
We look at the differences between all n pairs:

If the population means are the same (Null Hypothesis):
The statistic tstat follows a t-distribution (i.e. it should be close to 0!)
Paired vs Unpaired t-test

Φ = n1 + n2 – 2

Φ = n – 1

Vector of paired differences:
Paired t-test

Sample 1
Sample 2

Student’s t test (2-tailed)

There is only a 5% probability of finding
tstat higher than this value purely by chance…
There is only a 5% probability of finding
tstat higher than this value purely by chance…
There is
a 90% chance
of finding
tstat in this
range by chance
is tstat statistically distinguishable from zero? Example (90% conf)
tcrit
– tcrit

Two-tailed t-test

Null-Hypothesis: (H0): μ1 = μ2
Population means are not statistically different

Alternative Hypothesis: (H1): μ1≠ μ2
Population means are statistically different
two-tailed vs one-tailed t-test
One-tailed t-test

Null-Hypothesis: (H0): μ1 ≤ μ2
μ1 is not statistically significantly greater than μ2 (use H0: μ1= μ2 to test H1)

Alternative Hypothesis: (H1): μ1> μ2
μ1 is statistically significantly greater than μ2

two-tailed vs one-tailed t-test
Distribution of tstat TWO-tailed test

0
1- α
α/2
α/2
-tcrit
+tcrit
tcrit = – tinv(alpha/2,phi)

OR

tcrit = tinv(alpha/2+1-alpha,phi)

Distribution of tstat ONE-tailed test

0
1- α
α
tcrit
tcrit = tinv(1-alpha,phi)

OR (depending on H0)
0
1- α
α
tcrit
tcrit = tinv(alpha,phi)

Is it a paired (same population) or unpaired test ?
Is it a one-tailed (e.g. μ1> μ2) or two-tailed test (e.g. μ1 ≠ μ2) ?

Decide upon a level of significance α.
e.g. 99% and 95% are typical (α = 0.01 or 0.05)
Find tcrit using tinv (one- vs. two-tailed)
Find tstat from your sample (paired vs. unpaired)
Compare tstat and tcrit
If |tstat| > tcrit: the difference is significant (you can reject H0)*
else: the difference is not significant (you cannot reject H0)
Optional: determine the p-value (using tcdf of your tstat)

*This example if for a two-tailed test.
Summary

We are interested in ocean acidification. We measure the pH of ocean water at the pier of Newport Beach at two different dates:

In 1994: 8.03, 8.08, 7.99, 8.00, 7.93, 7.98
In 2004: 7.99, 8.02, 7.92, 7.94, 8.01, 7.93

From our two sample, we have:
In 1994: m1 = 8.0017 and s1 = 0.0504
In 2004: m2 = 7.9683 and s2 = 0.0406

Does the difference between the two means show a significant decrease or is it likely caused just by chance?

Example

What kind of t-test should you perform?

Paired t-test
Unpaired t-test
I don’t know

i>Clicker question

What kind of t-test should you perform?

One-tailed t-test
Two-tailed t-test
I don’t know

i>Clicker question

Choose a level of significance α = 0.1 (CI 90%)
This is a “one tailed” test (H1: m2 < m1) Numbers of degrees of freedom: Φ = 6 -1 = 5 tcrit = tinv(1-0.1,5) = 1.4759 Now, we have our critical value, what is our statistics? d = [8.03, 8.08, 7.99, 8.00, 7.93, 7.98] - [7.99, 8.02, 7.92, 7.94, 8.01, 7.93]; tstat = mean(d)/(std(d)/sqrt(length(d))); = 1.4464 tstat < tcrit : We cannot reject H0 Optional: p-value = 1-tcdf(tstat,5) = 0.1039 Example χ2-test “Goodness of fit” We want to compare an observed frequency distribution to a theoretical distribution Ex: we want to show that the yearly averaged rainfall in Irvine follows a normal distribution Ex: we want to make sure that a dice is not loaded χ2-test “Goodness of fit” We decompose the number of observations (n) over k intervals (or bins, or classes) k must satisfy n/k ≥ 5 k ≥ 10 So n ≥ 50 The Expected number of counts in any cell is Ei The Observed number of counts is Oi χ2 statistic χ2stat measures the mismatch between the Expected and the Observed distributions χ2stat = 0 perfect fit χ2stat large: poor fit χ2 statistic Our statistic χ2stat follows a χ2 -distribution ! Formulate a null and alternative hypothesis: H0: The data are consistent with a specified distribution H1: The data are not consistent with a specified distribution Choose a Significance level: α = 0.05 (5%) use MATLAB’S chi2inv function to find χ2crit Analyze Sample data Degrees of freedom = k-1 Calculate the expected frequency counts Ei Calculate the test statistic Interpret the results Conducting a χ2 test Is our dice loaded? Compare to a uniform distribution For alpha = 0.02: chi2crit = chi2inv(1-0.02,6-1) =13.3882 The die is loaded (98% confidence) Example Value Observed freq. Expected freq. (O-E)^2/E 1 16 10 3.6 2 5 10 2.5 3 9 10 0.1 4 7 10 0.9 5 6 10 1.6 6 17 10 4.9 Total 60 60 13.6 Lab 6: Hypothesis testing Lecture 7: Curve Fitting and interpolation Midterm Part 1 in class next Tuesday Midterm Part 2 in lab Wednesday next week What’s next? x̄ q N ! 1 P (x1 < x < x2) = Z x2 x1 f(x)dx µ X̄ = µ � X̄ = � p n µ X̄ = µ � x̄ = � p n µ X̄ = µ = 2L � X̄ = � p N = 0.7 p 50 P (x̄ > 2.2L)

� = 0.7L

µ = 2L

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AAACEHicbVC7TsNAEDzzDOFloKSxiBA0BDsgQRkBBWWQCCDFUbS+nMOJ89ncrRGR5U+g4VdoKECIlpKOv+EcXPAaaaXRzK52d4JEcI2u+2GNjU9MTk1XZqqzc/MLi/bS8pmOU0VZm8YiVhcBaCa4ZG3kKNhFohhEgWDnwdVh4Z/fMKV5LE9xmLBuBAPJQ04BjdSzN7b8UAHN/CMmEPwAVHab55mv+SCCbV9fK8xknld7ds2tuyM4f4lXkhop0erZ734/pmnEJFIBWnc8N8FuBgo5FSyv+qlmCdArGLCOoRIiprvZ6KHcWTdK3wljZUqiM1K/T2QQaT2MAtMZAV7q314h/ud1Ugz3uxmXSYpM0q9FYSocjJ0iHafPFaMohoYAVdzc6tBLMAGhybAIwfv98l9y1qh7O/XGyW6teVDGUSGrZI1sEo/skSY5Ji3SJpTckQfyRJ6te+vRerFev1rHrHJmhfyA9fYJ5HGdww==

�x̄

�/
p
n

AAACD3icbVBNS8NAEN34bf2KevQSLIqnmlRBj6IePCrYVmhCmWw37eJmE3cnYgn5B178K148KOLVqzf/jduag1YfDDzem2FmXpgKrtF1P62Jyanpmdm5+crC4tLyir261tRJpihr0EQk6ioEzQSXrIEcBbtKFYM4FKwVXp8M/dYtU5on8hIHKQti6EkecQpopI697UcKaO6fMoHgh6Dyu6LIfc17Mez6+kZhLoui0rGrbs0dwflLvJJUSYnzjv3hdxOaxUwiFaB123NTDHJQyKlgRcXPNEuBXkOPtQ2VEDMd5KN/CmfLKF0nSpQpic5I/TmRQ6z1IA5NZwzY1+PeUPzPa2cYHQY5l2mGTNLvRVEmHEycYThOlytGUQwMAaq4udWhfTABoYlwGII3/vJf0qzXvL1a/WK/enRcxjFHNsgm2SEeOSBH5Iyckwah5J48kmfyYj1YT9ar9fbdOmGVM+vkF6z3L3RCnYw=

µX̄
AAAB9HicbVBNS8NAEJ34WetX1aOXxSJ4KkkV9Fj04rGC/YAmlM120y7d3cTdTaGE/A4vHhTx6o/x5r9x2+agrQ8GHu/NMDMvTDjTxnW/nbX1jc2t7dJOeXdv/+CwcnTc1nGqCG2RmMeqG2JNOZO0ZZjhtJsoikXIaScc3838zoQqzWL5aKYJDQQeShYxgo2VAl+k/cwPscq6ed6vVN2aOwdaJV5BqlCg2a98+YOYpIJKQzjWuue5iQkyrAwjnOZlP9U0wWSMh7RnqcSC6iCbH52jc6sMUBQrW9Kgufp7IsNC66kIbafAZqSXvZn4n9dLTXQTZEwmqaGSLBZFKUcmRrME0IApSgyfWoKJYvZWREZYYWJsTmUbgrf88ipp12veZa3+cFVt3BZxlOAUzuACPLiGBtxDE1pA4Ame4RXenInz4rw7H4vWNaeYOYE/cD5/AFQkknY=


AAAB/nicbVDLSgMxFM34rPU1Kq7cBIvgqsxUQZdFNy4r2Ad0hiGTpm1okhmSjFDCgL/ixoUibv0Od/6NmXYW2nrgwuGce7n3njhlVGnP+3ZWVtfWNzYrW9Xtnd29fffgsKOSTGLSxglLZC9GijAqSFtTzUgvlQTxmJFuPLkt/O4jkYom4kFPUxJyNBJ0SDHSVorc40DREUeRiUwQI2l6eZ7DauTWvLo3A1wmfklqoEQrcr+CQYIzToTGDCnV971UhwZJTTEjeTXIFEkRnqAR6VsqECcqNLPzc3hmlQEcJtKW0HCm/p4wiCs15bHt5EiP1aJXiP95/UwPr0NDRZppIvB80TBjUCewyAIOqCRYs6klCEtqb4V4jCTC2iZWhOAvvrxMOo26f1Fv3F/WmjdlHBVwAk7BOfDBFWiCO9ACbYCBAc/gFbw5T86L8+58zFtXnHLmCPyB8/kDPdiVpw==

�x̄

s/
p
n

�x̄

µ

�12

< x̄� µ X̄ � X̄ < +12 � X̄ �12/� X̄ 12/� X̄ Z̄ = X̄ � µ X̄ � X̄ = X̄ � µ X̄ sp n x̄ = 2.3429 s = 1.0422 Z̄ = X̄ � µ X̄ � X̄ ' X̄ � µ X̄ sp n �x̄ s/ p n = 2.4469 �x̄ = 2.4469 s p n �x̄ = 0.9639 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

µ = 7



=


p
n

s
p
n

x̄ = 7.2

(x̄d, sd)

tstat =
m1 �m2q

s21
n1

+
s22
n2

tstat =
x̄d

sd/
p
n

2

6666666
4

a1
a2
a3

an�1
an

3

7777777
5

2

6666666
4

b1
b2
b3

bn�1
bn

3

7777777
5

d =

2

6666666
4

a1 � b1
a2 � b2
a3 � b3


an�1 � bn�1

an � bn

3

7777777
5

/private/tmp/tp16b596be_bb00_4e4f_9701_f9785f96c170.eps

Critical Value
-4 -3 -2 -1 0 1 2 3 4

D
e
n
s
it
y

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

�2stat =
kX

i=1

(Oi � Ei)
2

Ei