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Equations did not copy/paste right so do not refer to this page to do the assignment, as you know of course. 🙂 Refer to attached document/pdf.
For the project you should identify three research questions that can be addressed through different hypothesis testing procedures. Your submission should include the following components.
1.
Introduction: Briefly describe (in words) each of the three research questions and your motivation for studying them (not more than two pages). For each test you should state your null hypothesis and alternative hypothesis.
2.
Hypothesis tests: for each of the three tests clearly discuss: (1) the data used to conduct the test, (2) any assumption that you need to make in order to conduct the test including support for making these assumptions, (3) the test and its results, (4) the statistical significance of your findings.
3.
Summary: summarize your findings form the analysis. What do you conclude about the hypothesis you raised? Identify key limitations of your analysis (including any limitations of the data set), contributions that you believe your analysis makes to the readers, and any interesting directions for future analysis (not more than three pages)
Project sample is shown as below:
Introduction: The Doll Computer Company makes its own computers and delivers them directly to customers who order them via the Internet.
To achieve its objective of speed, Doll makes each of its five most popular computers and transports them to warehouses from which it generally takes 1 day to deliver a computer to the customer.
This strategy requires high levels of inventory that add considerably to the cost.
To lower these costs the operations manager wants to use an inventory model. He notes demand during lead time is normally distributed and he needs to know the mean to compute the optimum inventory level.
He observes 25 lead time periods and records the demand during each period.
The manager would like a 95% confidence interval estimate of the mean demand during lead time. Assume that the manager knows that the standard deviation is 75 computers
Hypothesis Tests:
25 observed lead time shown below
235 374 309 499 253
421 361 514 462 369
394 439 348 344 330
261 374 302 466 535
386 316 296 332 334
Interpret the question
·
X represents demand
a.
X~N (µ,75)
· δx=75 (standard deviation)
· n=25
· We would like to create a 95% confidence interval for the mean demand based on our sample of 25 lead time periods
· What do we need to know?
x̄
+Zα/2
· Compute the sample mean (X-bar) = 370.16
· Finding
Zα/2
a. Looking for
Zα/2
such that:
b.
P(-Zα/2
c.
Easiest:
Look for the lower tail probability in the normal table
P(Z<-Z
α/2
)=0.025
d.
Therefore:
Zα/2=
1.96
The computation is not done, but the format is pretty much like shown above.
For the project you should identify
three research
questions that can be addressed through different hypothesis testing procedures. Your submission should include the following components.
1. Introduction: Briefly describe (in words) each of the three research questions and your motivation for studying them (not more than two pages). For each test you should state your null hypothesis and alternative hypothesis.
2. Hypothesis tests: for each of the three tests clearly discuss: (1) the data used to conduct the test, (2) any assumption that you need to make in order to conduct the test including support for making these assumptions, (3) the test and its results, (4) the statistical significance of your findings.
3. Summary: summarize your findings form the analysis. What do you conclude about the hypothesis you raised? Identify key limitations of your analysis (including any limitations of the data set), contributions that you believe your analysis makes to the readers, and any interesting directions for future analysis (not more than three pages)
Project sample is shown as below:
Introduction:
The Doll Computer Company makes its own computers and delivers them directly to customers who order them via the Internet.
To achieve its objective of speed, Doll makes each of its five most popular computers and transports them to warehouses from which it generally takes 1 day to deliver a computer to the customer.
This strategy requires high levels of inventory that add considerably to the cost.
To lower these costs the operations manager wants to use an inventory model. He notes demand during lead time is normally distributed and he needs to know the mean to compute the optimum inventory level.
He observes 25 lead time periods and records the demand during each period.
The manager would like a 95% confidence interval estimate of the mean demand during lead time. Assume that the manager knows that the standard deviation is 75 computers
Hypothesis Tests:
25 observed lead time shown below
235 374 309 499 253
421 361 514 462 369
394 439 348 344 330
261 374 302 466 535
386 316 296 332 334
Interpret the question
· X represents demand
a. X~N (µ,75)
· δx=75 (standard deviation)
· n=25
· We would like to create a 95% confidence interval for the mean demand based on our sample of 25 lead time periods
· What do we need to know?
x̄ + Zα/2
· Compute the sample mean (X-bar) = 370.16
· Finding Zα/2
a. Looking for Zα/2 such that:
b. P(-Zα/2 c. Easiest: Look for the lower tail probability in the normal table P(Z<-Z α/2)=0.025 d. Therefore: Zα/2=1.96 The computation is not done, but the format is pretty much like shown above. INTRODUCTION TO HYPOTHESIS TESTING 9 a n d 1 1
D evo, 20 13 1 INFERENTIAL STATISTICS
¢ We often use statistics to test theories. — Theory: a prediction, or a group of predictions, about how
people, physical entities, and built devices behave.
¢ Theories begin as predictions, which are then the drawing of conclusions about a population of interest ¢ We will cover: HYPOTHESES TESTING
¢ Statistics is very much about expectations. We test.
¢ We study the probability of our sample’s outcome We believe that Our sample Possible?
X~N(10,2) 12 Probably yes
X~N(10,2) 20 Probably not HYPOTHESES
¢ We differentiate between the research hypothesis null hypothesis.
— Example: A bank manager argues that, on average,
people carry $ 50 or more in their wallet. This claim — where H0 is used to notate the null hypothesis and H1 HYPOTHESES ¢ Researchers tell you that, on average, people have ¢ A statistics professor wants to know if her section’s THREE FORMS OF HYPOTHESES
H0: Average H1: Average Lower- H0: Section H1: Section Two- number of H1: Average Upper- -4 -3 -2 -1 0 1 2 3 4
Z
The ‘equal’ sign (=, ≥, ≤) always goes in the null hypothesis A FINAL NOTE ON HYPOTHESES
¢ Hypotheses must always be (1) exhaustive and (2) possible option.
— mutually exclusive means there is no overlap between Not Exhaustive Exhaustive
H0: Average weeks of job seeking = 20 H0: Average weeks of job seeking ≤ 20 WRITE THE HYPOTHESES
¢ Angry Birds is one of the most popular games on D
evo, 2013
8 HYPOTHESES H0: Average time playing (µ) ≥ 10
H1: Average time playing (µ) < 10
D MORE PRACTICE
D ¢ Using the table as your guide number of sleep hours average hours spent grooming hours spent on educational — A two-tail test for the average — A two-tail test for the average — An upper-tail test for the Activity Hours
Sleeping 8.4
Leisure and Sports 3.6
Educational Activities 3.4
Working 3.0
Other 2.2
Traveling 1.5
Eating and Drinking 1.1
Grooming 0.8
Total 24.0 EXAMPLE
¢ Suppose that we are considering an investment in a ¢ Based on historical data provided by the current ¢ We are concerned that sales may have decreased over ¢ With µ being the hypothesized average spending (the D 11 WE TAKE A SAMPLE
¢ Data from twenty random customers shows that hypothesized mean, but we can attribute it to — Should we reject the null hypothesis based on this DECISION RULE
¢ The question we need to ask ourselves is: size n=20 out of a population with a mean of $100 and — If it is reasonably possible, then we have no reason to — If the chances of finding this sample mean are ¢ So, just how possible is it that we could obtain a D THE SAMPLING DISTRIBUTION OF D 14
To answer this question we need to know: SAMPLING DISTRIBUTIONS
¢ A sampling distribution refers to the ¢ Let us continue with our used textbook store D 15 SAMPLING DISTRIBUTION
¢ Recall that we took a random sample of twenty ¢ Let us repeat this sampling procedure 200 times, D 16
Sample (n=20) Sample Mean (X) 27
2 $104. 34 3 $100. 49
… …
199 $96. 65
200 $111.65
Average over 200 samples: $99.04 GRAPHICALLY
D 17
F q e n
cy
F 200 samples of size 20
1,000 samples of size 20 CENTRAL LIMIT THEOREM (PART I)
D 18
¢ The central limit theorem states that the ¢ This standard deviation ( ) is called the σ
n
σ X ~ N(100, 35 BUT WHAT ABOUT NON-NORMAL ¢ Consider this dice-rolling example: one of the values from ‘1’ to ‘6’, each with a D 19 0 0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
1 2 3 4 5 6 ¢ Instead of rolling a single die, we will roll two ‘5’, then we will record the outcome of this roll as ‘4’.
¢ Exercise: obtain along with their probabilities
D GRAPHICALLY 21 0.020
0.0 40 0.0 60 0.080
0.100
0.120
0.140
0.160
0.180
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 AS WE INCREASE THE NUMBER OF DICE D 22 0.01
0.02 0.03
0.04 0.05
0.06 0.07
0.08 CENTRAL LIMIT THEOREM (PART II)
D 23
¢ even when the population is not normally n PUTTING IT ALL TOGETHER
D 24
The mean of a sample taken from a normally The mean of a sufficiently large sample taken from a Using statistical notaion: EXAMPLE 9.1(A)
¢ The foreman of a bottling plant has observed that that the bottle will contain more than 32 ounces?
¢ Answer: D 25 7486.2514.1)67.Z(P ⎞ ⎛ − σ µ− EXAMPLE 9.1(B)
¢ If a customer buys a carton of four bottles, 32
ounces? asked about the mean of X
D 26
P(X > 32) = ? INTERPRET THE QUESTION
¢ We know that ¢ Therefore:
D €
X ~ N(32.2, 0.3 )
€
X ~ N(µ, σ SOLUTION
¢ Answer: There is about a 91% chance that the D 28
€ Z = X − µX = = −1.33
P(X > 32) = P(Z > −1.33) = 0.9082 BACK TO HYPOTHESIS TESTING
D 29 EXAMPLE ¢ The Freshman 15 refers to the weight gained (in D 30
€ H0 : µ ≥15 lbs [or µ =15] TEST STATISTIC
¢ The first thing we need to do is to convert our ¢ In our example:
D 31
Z = X −µ n Z = 13−15 = −2 P-VALUE
¢ The probability of finding a sample mean as ¢ Our decision rule for whether or not to reject the D P(X <13)→ P(Z < 13−15
10 100
) = P(Z < −2) = 0.0228 THE SIGNIFICANCE OF THE TEST
¢ We compare the p-value to a desired significance ¢ In statistics, when we say that a finding is ¢ Hence, we form the following decision rule for our null hypothesis. 33 IN OUR EXAMPLE
¢ If we choose α=0.05 then: ¢ Therefore we reject the null hypothesis and D ANOTHER EXAMPLE (11.34)
¢ Spam email has become a serious and costly D 35 48 29 44 17 21 32 28 34
23 13 9 11 30 42 37 43 48 INTERPRETATION office workers reading and deleting spam exceeds 25 minutes per day” ¢ “a random sample of 18 workers” ¢ “measures the amount of time each spends reading and deleting spam
¢ “The results are listed below.” ¢ “If the population of times is normal with a standard deviation of 12 ¢ “can the manager infer at the 1% significance level that he is correct” D 36 STEP 1: COMPUTE THE TEST STATISTIC
¢ The test statistic is the Z-score corresponding to ¢ We know that X~N(25,12) and therefore, based ¢ Test statistic:
D 37
X ~ N(25, 12 Z = 30.22− 2512 =1.845 STEP 2: FIND THE P-VALUE
¢ This is an upper level test sample mean at least as high as the one we have — P(Z > 1.85) = 1 – 0.9678 = 0.0322 D 38 STEP 3: DECISION RULE AND CONCLUSION
¢ We will reject the null hypothesis if p-value is less than α. ¢ By not rejecting the null hypothesis we conclude that the
¢ But what if we chose α to be 5%?
— In this case p-value < α and we reverse our conclusion and
reject the null hypothesis
¢ It is not uncommon to change our conclusion as α changes. of our test. level but not at the 1% level.
D 39 IN RESEARCH PAPERS
D A TWO-TAIL TEST EXAMPLE
¢ A statistics professor wishes to determine if her D 41 STEP 1: HYPOTHESES
¢ This is a two tail test in which we would reject D 42
H0 :µ = 75 STEP 2: COMPUTE THE TEST STATISTIC
¢ Let X represent statistics grades. that X is normally distributed with a mean of 75 and — We also know that the sampling distribution of X-bar ¢ To obtain the test statistic, we convert our 72 to a Z score using the D 43
X ~ N(75, 4 Z = 72− 75 = −3.75 STEP 3: P-VALUE
¢ What is the probability of finding a test statistics ¢ Because this is a two-tail test p-value needs to be 0.00018
D STEP 4: DECISION RULE AND CONCLUSION
¢ Because our p-value of ‘0.00018’ is less than α (0.05) we ¢ At α=0.05 we can reject the null hypothesis and D 45 KEY TERMS
¢ A Hypothesis is a claim we wish to test by taking repeated samples of size n from a given sampling distribution of the mean in most cases (see ¢ Test statistic is the value we use to conduct our test. ¢ P-value is the probability of finding a test statistic as D 46 ANOTHER APPROACH (SIMILAR BUT NOT ¢ A baseball player claims that the average speed D 47 STEP 1: HYPOTHESES ¢ The pitcher would like to prove that his fastball ¢ with µ being the hypothesized average pitch D STEP 2: COMPUTE THE TEST STATISTIC ¢ Let X represent pitching speed (in mph). that X is normally distributed with a mean of 90mph — We also know that the sampling distribution of X-bar ¢ To obtain the test statistic, we convert our D X ~ N(90, 3.85 ) Z = 90.8− 90 = 2.078 STEP 3: FIND THE CRITICAL VALUE
¢ To determine the critical value, we need to know at the 1% level of significance: α=0.01. P(Z ≥ |critical value|) = α. find the critical value Z*, such that: P(Z ≥ Z*) = 0.01.
¢ Note that we are using the ‘greater than’ distribution.
¢ You’ll see that our critical value Z* is 2.326.
D GRAPHICALLY ¢ The shaded blue area is called the rejection reject the null hypothesis 51 STEP 4: COMPARE THE TWO VALUES
¢ Our test statistic does not fall within the D 52 STEP 5: CONCLUSION
¢ We were not able to reject the null hypotheses D 53 A LOWER TAIL TEST EXAMPLE
¢ The director of a state agency claims that the deviation is known to be $2,500 and the significance Dorit Nevo, 2013 STEP 1: HYPOTHESES ¢ This is a lower tail test in which we would only D 55
€ H0 :µ ≥ $30,000 Dorit Nevo, 2013 -4 -3 -2 -1 0 1 2 3 4 ONE-TAIL TEST (LOWER TAIL)
Lower tail rejection STEP 2: COMPUTE THE TEST STATISTIC ¢ Let X represent starting salaries ($). that X is normally distributed with a mean of $30,000 — We also know that the sampling distribution of X-bar ¢ To obtain the test statistic, we convert our D 57
X ~ N(30000, 2500 ) Z = 29570−30000 = −1.72 STEP 3: CRITICAL VALUE
¢ The question defines the desired significance ¢ Looking for Z critical such that P(Z<-Z*)=0.05
(lower rejection region only)
¢ From the table we can find that Z* = -1.645
D 58 STEP 4: COMPARE THE TWO VALUES 59
€ Zstatistic = −1.72 < −1.645 = Zcritical STEP 5: CONCLUSION ¢ The computed Z statistics is more extreme than our D A TWO-TAIL TEST EXAMPLE 61 STEP 1: HYPOTHESES ¢ This is a two tail test in which we would reject D 62
H0 :µ = 75 STEP 2: COMPUTE THE TEST STATISTIC 63
X ~ N(75, 4 STEP 3: CRITICAL VALUE ¢ Because this is a two tail test we should have two ¢ This means that α should be split in half for each of lower tail) D 64 STEP 4: COMPARE THE TWO VALUES Zstatistic = −3.75< −1.96 = Zcritical STEP 5: CONCLUSION ¢ The computed Z statistics is more extreme than our ¢ At α=0.05 we can reject the null hypothesis and 66 TYPE I AND TYPE II ERRORS
Still Chapter 11
D 67 STARTING WITH AN EXAMPLE
¢ It has been claimed the average wireless phone ¢ Step 1: Hypotheses — This is an upper tail test, α is 0.05, therefore our critical D 68 CONT’D
D 69
¢ Step 3: test statistic
¢ Steps 4 and 5: the test statistic to our critical value of 1.645 and Z = 8.2−8 =1.94 TYPE I AND TYPE II ERRORS ¢ When testing hypotheses we can make two kinds ¢ Concluding that the mean number of spam messages is ¢ This is called a type I error
— Not reject a false null hypothesis more than eight when in fact it is. Dorit Nevo, 2013 ERROR SUMMARY
Hypothesis testing In actuality
H0 is true H0 is false Not reject H0
Reject H0
Dorit Nevo, 2013 Type I Type II TYPE I ERROR IN OUR EXAMPLE
¢ We assumed the null hypothesis to be true and 60 with a mean of 8.2 spam message is: P(Z ≥ 1.94)=0.026, — Based on our significance level, α, we concluded that we ¢ Assume that the null hypothesis is, in fact, true. type I error. significance), α is also the probability of committing a type D THE TRADE-OFF
¢ If we wish to reduce the chance of committing a ¢ Why not use the lowest α possible? committing a type II error (not rejecting a false null — This probability is called β, and it is inversely related D evo, 2010
73 SUMMARY ¢ We covered Chapters 11 as well as section 9.1 ¢ Next class – Quiz 12 (other single population tests)
D 74
Chapters
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repeatedly tested in various settings to either
strengthen or refute them.
— Such testing often involves statistical inference, defined as
based on findings from samples obtained from that
population.
— Hypothesis testing
— Analysis of Variance (ANOVA)
— Regression Analysis
aim to test specific expectations that we have
about the population’s parameters using sample
statistics. We call these expectations hypotheses.
— A hypothesis is some specific claim that we wish to
given the hypothesized distribution of the
population
mean is
and the
is the null hypothesis. The research hypothesis
contains the other side of this claim, that is – that
people carry less than $50. We can also write it as:
¢ H0: Average amount of money ≥ $50
¢ H1: Average amount of money < $50
(sometimes denoted HA) is used to notate the
research hypothesis, commonly referred to as the
alternative hypothesis.
200 or fewer friends on Facebook. However, you
believe that Facebook users, in fact, have more than
200 friends. You can set your hypotheses as:
— H0: Average number of friends ≤ 200
— H1: Average number of friends > 200
grade average is different than that of other sections
of the course. The average for all other sections is
‘75’. To help the professor learn if her section’s grade
average is different than that of other sections, we
need to set up the following hypotheses:
— H0: Section’s grade average = 75
— H1: Section’s grade average ≠ 75
amount of money
≥ $50
amount of money
< $50
Tail
Test
average = 75
average ≠ 75
Tail
Test H0: Average
friends ≤ 200
number of
friends > 200
Tail
Test
mutually exclusive.
— Exhaustive means that the hypotheses should cover every
hypotheses. The following hypotheses are, therefore,
incorrect because the ‘= 20’ option is in both the null and
alternative hypotheses:
¢ H0: Average weeks of job seeking ≥ 20
¢ H1: Average weeks of job seeking ≤ 20
H1: Average weeks of job seeking > 20
H1: Average weeks of job seeking > 20
various platforms. According to one source, people
collectively spend 300 million minutes per day
playing Angry Birds and it supposedly costs the US
economy billions of dollars in lost work time. Suppose
you believe this number (300 million minutes) is too
high. You conduct some more research to find out
that there are roughly 30 million daily active users of
Angry Birds, which means that on average each
player spends (300 million minutes/30 million users)
10 minutes per day playing Angry Birds, according to
the original claim. You plan to use a sample of
students at your school to test this claim and need to
set up your hypotheses first.
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9
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10
for the three types of tests
(lower-tail, upper-tail, and two-
tail), write the hypotheses for
the following tests:
— A two-tail test for the average
— An upper-tail test for the
— A lower-tail test for the average
activities
hours spent eating and
drinking
hours spent on leisure and
sports
average hours spent working
used textbook store and are interested in determining
the amount of money that customers are likely to
spend when shopping at the store.
owner, the average customer making purchases at the
store spent $100 with a standard deviation of $35.
the last year. We hypothesize that:
— H0: Average spending (µ) ≥ 100
— H1: Average spending (µ) < 100
parameter of interest) for the population (all
customers who have made purchases at the store).
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the mean spending in our sample is ‘$93.27’.
— Our sample mean is indeed lower than the
sampling error.
one sample?
— how possible is it for us to take a random sample of
a standard deviation of $35 and obtain a sample
mean of ‘$93.27’?
reject the null hypothesis.
extremely low, then we should conclude that the null
hypothesis is likely false and we should reject it,
believing that the true average spending is, in fact,
lower than $100.
sample mean of ‘$93.27’ given a population mean
of $100 and standard deviation of $35?
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13
THE MEAN
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probability distribution of any sample statistic.
The sampling distribution of the mean
describes the probabilities attached to all values
of the mean of samples that are repeatedly taken
from the same population.
example to better understand this concept.
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customers (i.e., n=20), and measured how much
each of the customers spent. We then computed
the mean spending for that sample to be ‘$93.27’.
each time taking a different sample of twenty
customers and computing the mean spending for
each sample.
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1 $93.
Standard Deviation: $11.14
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re
u
re
q
u
en
cy
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sampling distribution of the mean of a random
sample of any size (n) drawn from a normally
distributed population also follows a normal
distribution with a mean of µ and a standard
deviation of .
standard error of the mean.
— In our example we thus know that:
n
20
)
POPULATIONS?
— When we roll a single six-sided die, we can obtain any
probability of ‘1/6’. Defining a random variable as
‘the outcome of a single die roll’, this variable follows
the uniform and discrete distribution.
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dice. And, instead of looking at the outcome on
the individual die, we will define our random
variable as ‘the average of the values obtained
from two rolled dice’.
— For example, if one die rolled on ‘3’ and the other on
— List all possible values that the random variable can
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20
D
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0.000
ROLLED (10 IN THE GRAPH BELOW)…
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0
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
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distributed, the mean will follow an
approximately normal distribution for large
enough samples (n≥30), with a mean of µ and a
standard deviation of . σ
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distributed population (e.g., the spending example)
will follow a normal distribution, with the mean equal
to the population mean and a standard deviation
equal to the population standard deviation divided by
the square root of the sample size (hence the
distribution is narrower).
non-normally distributed population (e.g., the dice
example) will follow an approximately normal
distribution, with the mean equal to the population
mean and a standard deviation equal to the
population standard deviation divided by the square
root of the sample size.
the amount of soda in each “32-ounce” bottle is a
normally distributed random variable, with a
mean of 32.2 ounces and a standard deviation of
0.3 ounce.
— If a customer buys one bottle, what is the probability
— We know that X~N(32.2,0.3)
— Using the standard normal table:
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3.
2.3232XP)32X(P =−=−>=⎟
⎠
⎜
⎝
>
=>
what is the probability that the mean amount
of the four bottles will be greater than
— We are no longer asked about X. Instead we are
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— X ~ N(32.2 , 0.3)
— n=4 (a carton of four bottles)
— The central limit theorem à
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27
4
n
)
mean of the four bottles contains more than 32oz.
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σX / n
32 − 32.2
0.3/2
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lbs) during a student’s first year at college. A
researcher set out to challenge this belief
claiming that the weight gain is in fact much
lower. Based on a sample of 100 students the
researcher found an average weight gain of 13lbs.
Assuming the standard deviation of weight gain
is known to be 10lbs, what can the researcher
conclude?
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HA : µ <15 lbs
sample mean into a Z-score. This Z-score is
known as the test statistic
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σ
10 / 100
extreme as the one we have found is called the p-
value of the test.
null hypothesis will be based on this probability.
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32
level known as α (alpha).
‘statistically significant’, it means that the
finding is unlikely to have occurred by chance.
Our level of significance is the maximum
chance probability we are willing to tolerate.
hypothesis testing context:
— For any p-value smaller than α we reject the
D
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P-value = 0.028 < 0.05 = α
conclude that the average weight gain is likely
lower than 15lbs.
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34
nuisance. An office manager believes that the
average amount of time spent by office workers
reading and deleting spam exceeds 25 minutes
per day. To test this belief, he takes a random
sample of 18 workers and measures the amount
of time each spends reading and deleting spam.
The results are listed below. If the population of
times is normal with a standard deviation of 12
minutes, can the manager infer at the 1%
significance level that he is correct?
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35
¢ “An office manager believes that the average amount of time spent by
— H0: µ ≤ 25
— H1: µ > 25
— n=18
spam”
— X is a random variable defined as the time spent reading and deleting
— We can compute the sample mean as 30.22
minutes”
— The hypothesized distribution is X~N(25,12)
— α=0.01
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our sample mean
on the central limit theorem, we know that:
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18
)
18
— We are looking for the probability of obtaining a
found:
¢ I round Z to 1.85 to be able to look it up in the table
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— P-value = 0.0322 and the questions states that α is 0.01
— Since p-value > α we do not reject the null hypothesis.
time spent on spam likely does not exceed 25 minutes.
— We use the p-value to indicate the highest level of significance
— In this above example, our test would be significant at the 5%
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40
section’s grade average is different than that of
other sections of the course. The average for all
other sections is 75 with a standard deviation of
4. The professor collected test scores from 25
students and found the sample average was 72.
She wishes to conduct a hypothesis test at the 5%
level of significance (that is, α=0.05).
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the null hypothesis if the sample mean is
significantly smaller or significantly larger than
the hypothesized mean.
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HA :µ ≠ 75
— Assuming the null hypothesis to be true, we believe
a standard deviation of 4.
(the average starting salary) is:
sample mean of
hypothesized distribution:
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25
)
4 25
as extreme as the one we have found?
— P( Z < -3.75) = 0.00009
doubled before we compare it to α
— p-value = 2*P(Z≤|test statistic|) = 2*0.00009 =
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44
reject the null hypothesis
conclude that the average grade is different than 75.
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¢ The sampling distribution of the mean is created
population, computing the mean for each sample, and
charting the distribution of these means.
— We rely on the central limit theorem to know the
chapter 9.1 for a review)
It is a Z-score calculated based on the sample statistic
and the hypothesized distribution
extreme (either as high or as low) as the one we have
found
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EXACTLY THE SAME)
of his fastball pitch is greater than that of his
rival, who averages 90 mph. He collects data on
the speed of 100 fastballs and finds his average
speed is 90.8 mph. Assuming that we know the
standard deviation of a fastball pitch to be 3.85
mph, what can we conclude about the pitcher’s
claim?
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has a higher speed than that of his rival.
Therefore, the null hypothesis is that it is not
higher (i.e. it is the same speed or lower) and the
alternative is that it is higher:
— H0: µ ≤ 90
— H1: µ > 90
speed.
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— Assuming the null hypothesis to be true, we believe
and a standard deviation of 3.85mph.
(the average pitching speed) is:
sample mean of 90.8 to a Z score using the
hypothesized distribution:
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100
3.85 100
the desired significance level (α).
— Let us assume that we would like to conduct the test
— The critical value is computed based on α, such that
— In our example: use the normal distribution table to
relationship here since this is an upper-tail test
— our critical value should be at the upper tail of the
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region.
— If our sample values fall within this region we would
D
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rejection region and therefore we do not reject
the null hypotheses.
— Test statistic 2.078 < 2.326 critical value
— Do not reject the null hypothesis
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concluding that there is no evidence to support
the claim that the pitcher’s fastballs are faster
than his rival’s
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average starting salary for clerical employees in
the state is less than $30,000 per year. To test
this claim, she has collected a random sample of
100 starting salaries of clerks from across the
state and found that the sample mean is $29,570.
— State the null and alternative hypotheses
— Conduct the test, assuming the population standard
level for the test is 0.05
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reject the null hypothesis if the test statistic is
significantly smaller than the hypothesized
parameter.
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HA :µ < $30,000
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Z
region
— Assuming the null hypothesis to be true, we believe
and a standard deviation of $2,500.
(the average starting salary) is:
sample mean of $29,570 to a Z score using the
hypothesized distribution:
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100
2500 100
level for the test as 0.05
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D
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critical Z value and therefore falls in the rejection
region. At α=0.05 we can reject the null hypothesis and
conclude that the average salary is less than 30,000.
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¢ A statistics professor wishes to determine if her
section’s grade average is different than that of
other sections of the course. The average for all
other sections is 75 with a standard deviation of
4. The professor collected test scores from 25
students and found the sample average was 72.
She wishes to conduct a hypothesis test at the 5%
level of significance (that is, α=0.05).
D
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the null hypothesis if the test statistic is
significantly smaller or significantly larger than
the hypothesized parameter.
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HA :µ ≠ 75
¢ Let X represent statistics grades.
— Assuming the null hypothesis to be true, we believe
that X is normally distributed with a mean of 75 and
a standard deviation of 4.
— We also know that the sampling distribution of X-bar
(the average starting salary) is:
¢ To obtain the test statistic, we convert our
sample mean of 72 to a Z score using the
hypothesized distribution:
D
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25
)
Z = 72− 75
4 25
= −3.75
rejection regions
— At both tails of the distribution
the tails:
— The question defines the desired significance level at 0.05
— Looking for Z critical such that P(Z<-Z*)=0.025 (this is the
— From the table we can find that Z* = -1.96
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D
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critical Z value at the lower tail, and therefore falls in
the rejection region.
conclude that the average grade is different than 75.
D
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customer receives no more than 8 spam text messages
each day, with a standard deviation of 0.8. To test
this claim (at the 5% significance level), you collect
data from 60 wireless customers and record the
number of spam messages received during a single
day. Your sample average is 8.2.
— H0: µ ≤ 8
— H1: µ > 8
¢ Step 2: critical value
value is 1.645
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— Testing at the 5% level of significance, we compare
conclude that the null hypothesis should be rejected,
because 1.94>1.645. In other words, we conclude that
the average customer receives more than 8 spam text
messages per day
0.8 60
of errors:
— Reject a true null hypothesis
more than 8 when in fact it is eight
¢ Concluding that the mean number of spam messages is not
¢ This is called a type II error
70
indicates that you
should
(“do not reject”)
(“reject”)
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error
(prob. α)
error
(prob. β)
computed the probability of finding the sample mean
that we observed.
— In our example the probability of finding a sample of size
which is quite low.
should reject the null hypothesis: we believe that it is false.
— By rejecting a true null hypothesis we have committed a
— Because our rejection rule is based on α (the level of
I error.
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type I error, we need to use smaller values of α.
— Because we also have to consider the probability of
hypothesis).
to α
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— Logic and process of hypothesis testing
¢ Next topics: chapter 10 (confidence intervals) and
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