Download Testing a Claim

Testing a
Claim
Chapter 11
1.1 Intro
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“I can make 80% of my BBall free throws”. To test my claim, you ask
me to shoot 20 free throws. I only make 8  and you say “aha!
Someone who makes 80% of their free throws would NEVER only
make 8/20!”
But I say, ‘hey, what if I’m having a bad day, or am injured, or the
ball is dead, or the hoop is bent…” Since all of these are possibilities,
you say “OK, well I’ll decide how likely your claim is based on the
probability that someone who genuinely makes 80% would shoot
8/20 on one trial run”

This is the basics of hypothesis testing: an outcome that
would rarely happen if a claim were true is good evidence
that the claim is not true.
What we will need for formal
Hypothesis Testing…
Stating a hypothesis
A
statistical test starts with a careful statement of
the claims we want to compare. Because the
reasoning of tests look for evidence against a
claim, we start with the claim we seek evidence
AGAINST.
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This is called our NULL HYPOTHESIS. (H0)
The claim about the population we are trying to find
evidence FOR is our ALTERNATIVE hypothesis (HA)
Example (one-sided)
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Attitudes towards school and study habits on a national
survey range from 0 to 200. The mean score for US
college students is 115 with a standard deviation of 30.
Assume normality in this population. A teacher suspects
that older students have better attitudes towards school.
She gives the survey to a SRS of 25 students.
 We seek evidence AGAINST the claim that μ = 115.
 Null: H0:
 Alternate: Ha:
 *Be sure to state the hypothesis in terms of population
b/c we are making inference/claims about our pop!
Two sided hypothesis

If the previous example said the teacher thought that
seniors had a DIFFERENT attitude towards study habits
(but didn’t specify better or worse), we would be
doing a two sided hypothesis because the alternate is
that μ could be > or < 115. So we state it as:
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H0:
Ha:
**The alternative hypothesis should express the hopes or
suspicions we have before we see the data. It is
cheating to first look at the data and then frame Ha to fit
what the data show!
Conditions for Significance tests
 The
same 3 conditions from Chapter 10 should be
verified before performing a significance test
about a population mean or proportion.
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1. SRS
2. Normality:
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Means (𝜎 known) -> CLT
Means (𝜎 unknown)  n<15, n ≥15, n ≥ 30
Proportions  np ≥ 10 and n(1 – p) ≥ 10
3. Independence: N≥10n and/or individual
observations are independent
Test Statistic
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
The significance test compares the value of the
parameter (true pop mean, as stated in the null) with the
calculated sample mean. Values of the sample far from
the true parameter give evidence against H0
To assess how far the sample statistic is from the
population parameter, we have to standardize it (to
make comparison)
estimate − hypothesized value
 Test statistic =
standard deviation of estimate
Z test

We will focus on the Z test first, which is when we
know sigma, so the Z test statistic formula is:
Attitudes towards school and study habits on a
national survey range from 0 to 200. The mean
score for US college students is 115 with a standard
deviation of 30. Assume normality in this
population. A teacher suspects that older students
have better attitudes towards school. She gives
the survey to a SRS of 25 students.


In that last example, lets say the mean (X bar) of
the 25 seniors sampled was 125. Our calculated Z
would be
Test statistic =
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A South Florida newspaper reports that the average cost of renting
a car in Fort Lauderdale during the months of January and February
is $36 per day. A random sample was taken of 40 people who
rented cars at Fort Lauderdale Airports and the mean of this sample
was found to be 𝑥= $34.20. The population standard deviation is
known to be 𝜎 = 2. We want to test the claim that the average
rental in Fort Lauderdale is less than that reported by the
newspaper.
Which of the following would be the correct test statistic?
(a) 0.0636
(b) -0.0636
(c) 0.127
(d) -0.045
P-Value
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We use our test statistic to calculate our p-value
The p value is the probability of getting your observed statistic,
assuming the null is true.
Small p-value = reject our Null Hypothesis
 The p value is the area under the curve from your calculated test
statistic, to the tail end.
For Example:

Test statistic = 1.67
Null: H0: μ= 115
Alternate: Ha: μ> 115
P-Value
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
The p value is the area under the curve from your
calculated test statistic, to the tail end.
Be careful if you are performing a two-sided test!!!
For Example:

Test statistic = 1.67
Null: H0: μ= 115
Alternate: Ha: μ ≠ 115
In a test of H0: p = 0.7 against Ha: p ≠ 0.7, a sample size of 80
produces z = 0.8 for the value of the test statistic. Which of the
following is closes to the P-value of the test?
Interpretation: Statistical Significance
 We
set a “maximum” p-value before calculating
our observed test statistic and we call this our
significance level.
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𝛼 is the symbol for our chosen significance level we
need to beat. So we would say 𝛼 =.05
Most commonly we choose 𝛼 =.05
 meaning
.05,
we need our calculated p value to be less than
If
we “beat” our alpha level –
Reject H0
If we do not “beat” our alpha
level -- fail to reject H0

If our calculated (or observed) p-value is less than or equal to our
alpha level, we say that the data is “statistically significant at level α”

When our data is statistically significant, we can reject our null
hypothesis.
Statistically Significant?
Reject H0?
One sided:
p-value = 0.047
Two sided:
p-value = 0.0949
𝜶 = 𝟎. 𝟎𝟏
𝜶 = 𝟎. 𝟎𝟓
𝜶 = 𝟎. 𝟏

“significant” doesn’t mean important! It just means you are
rejecting your null hypothesis
Inference Toolbox!
•To test a claim about an unknown population parameter:
•Step 1: State
•Identify the parameter (in context) and state your hypotheses
•Step 2: Plan
•Identify the appropriate inference procedure and verify the
conditions for using it (SRS, Normality, Independence)
•Step 3: Calculations
•Calculate the test statistic
•Find the p-value
•Step 4: Interpretation
•Interpret your results in CONTEXT
•Interpret P-value or make a decision about H0 using statistical
significance
Example =)
•Mel N. Colly is interested in whether or not his new treatment
for depressed patients is having any effect on his patients’
rating of depression. Suppose all of his depressed patients
have a mean depression score of 8 with a standard deviation
of 4. Mel chooses a random sample of 100 depressed
patients treated with his innovative approach and determines
that the mean depression score for these individuals is 7.5.
Does the cream have any effect?
•Mel N. Colly is interested in whether or not his new treatment
for depressed patients is having any effect on his patients’
rating of depression. Suppose all of his depressed patients
have a mean depression score of 8 with a standard deviation
of 4. Mel chooses a random sample of 30 depressed patients
treated with his innovative approach and determines that the
mean depression score for these individuals is 7.5. Does the
treatment have any effect?
•Step 1: STATE
•We are interested in 𝜇, the true mean
depression score of all of Mel’s
depressed patients
•H0 :
•HA :
•Mel N. Colly is interested in whether or not his new treatment for depressed patients is having
any effect on his patients’ rating of depression. Suppose all of his depressed patients have a
mean depression score of 8 with a standard deviation of 4. Mel chooses a random sample of 30
depressed patients treated with his innovative approach and determines that the mean depression
score for these individuals is 7.5. Does the treatment have any effect?
•Step 2: PLAN
•We will conduct a ______________________________.
•(1) SRS:
•The data was collected “at random.” The study does not state that
a simple random sample was used, but we will proceed assuming
proper sampling methods were used.
•(2) Normality:
•We do not know if the population distribution of depression
patients’ depression scores is Normal, but the sample size is large
enough (n=30) so that the sampling distribution will be
approximately normal (by the central limit theorem)
•(3) Independence:
•Mel N. Colly selected the patients without replacement, but we will
assume that there are more than 30(10) = 300 depressed patients
seen in his practice.
•Mel N. Colly is interested in whether or not his new treatment for depressed
patients is having any effect on his patients’ rating of depression. Suppose
all of his depressed patients have a mean depression score of 8 with a
standard deviation of 4. Mel chooses a random sample of 30 depressed
patients treated with his innovative approach and determines that the mean
depression score for these individuals is 7.5. Does the treatment have any
effect?
•Step 3: Calculations
•(1) Test Statistic
z = x-bar - μ0
σ/√n
•(2) P-value: Draw a picture using the standardized value, then calculate
the P-value
•Mel N. Colly is interested in whether or not his new treatment for depressed patients
is having any effect on his patients’ rating of depression. Suppose all of his depressed
patients have a mean depression score of 8 with a standard deviation of 4. Mel
chooses a random sample of 30 depressed patients treated with his innovative
approach and determines that the mean depression score for these individuals is 7.5.
Does the treatment have any effect?
•Step 4: Interpretation
•P-value (the problem did not give us an alpha level)
•A sample mean depression score of 7.5 would happen
49.36% of the time by chance if the true population mean
depression score was 8. Because the probability of
obtaining these results is so high, this is not good evidence
that the true mean depression score is not 8.
•Di Perrs is the quality control manager for Pampers. Pampers
claims the average absorbency of Pampers is 195 milliliters with a
standard deviation of 80 milliliters. A sample of 100 Pampers were
selected at random and tested. The average amount of fluid
absorbed was x-bar = 210 milliliters. Di Perrs wants to use an α =
0.05 significance level to determine whether or not the company’s
claim is true.
•Step 1: STATE
•We are interested in 𝜇, the true mean absorbency in
ml of all Pampers.
•H0 :
•HA :
•Step 2: PLAN
We will perform a 1-sample z-test for means (sigma known)
•(1) SRS: The data was collected “at random.” The study does not state
that a simple random sample was used, but we will proceed assuming
proper sampling methods were used.
•(2) Normality: We do not know if the population distribution of
Pampers absorbency is Normal, but the sample size is large enough
(n=100) so that the sampling distribution will be approximately normal (by
the central limit theorem)
•(3) Independence: Di Perrs selected the diapers without
replacement, but we can assume that there are more than 10(100) = 1000
diapers produced at the factory.
•Di Perrs is the quality control manager for Pampers. A recent ad claimed that the new improved
Pampers is more absorbent than the old Pampers. The average absorbency of old pampers was
195 milliliters with a standard deviation of 80 milliliters. A total of 100 new Pampers were selected
at random and tested. The average amount of fluid absorbed was x-bar = 210 milliliters. Di Perrs
wants to use an α = 0.05 significance level.
•Step 3: Calculations
•(1) Test Statistic
•(2) P-value: Draw a picture using the standardized value, then
calculate the P-value
•Di Perrs is the quality control manager for Pampers. A recent ad claimed
that the new improved Pampers is more absorbent than the old Pampers.
The average absorbency of old pampers was 195 milliliters with a standard
deviation of 80 milliliters. A total of 100 new Pampers were selected at
random and tested. The average amount of fluid absorbed was x-bar = 210
milliliters. Di Perrs wants to use an α = 0.05 significance level.
•Step 4: Interpretation
•Using significance Level
•Since our P-value,
La Tarea…
Prom
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Worried about his prospects for the prom, Malcolm claims
that girls at NPHS are a bit snobby and that the average
number of girls that a guy must ask to the prom before getting
a “yes” is 4.
Doug disagrees. He doesn’t think the girls are that snobby
and that the average number of girls that a guy must ask out
before getting a positive response is less than 4.
An SRS from last year of 50 junior and senior guys found that
the average number of girls that a guy asked out to the prom
was 3.4. Assuming the standard deviation from the entire
population is 𝜎 = 2, is there enough evidence to support
Alex’s claim (at the level 𝛼 = 0.05?)