Transcript Significance Test - Parkway School District

Significance Test
A claim is made.
Is the claim true?
Is the claim false?
• A test of signicance assesses the evidence
found in the data against a null hypothesis H o
•
in favor of an alternative hypothesis .H a
Null Hypothesis
• The null hypothesis H0 is always of the form:
• H0 :    o
The null hypothesis says that "the claim is true
and that what we observe in the
data happened by chance."
Alternative Hypothesis
• The alternative hypothesis Ha always takes one of three
forms:
•
Two-sided Ha :  
0
•
One-sided on the high side Ha :    o
•
One-sided on the low side Ha :
•
Note 1: The form of the alternative hypothesis is problem
dependent.
•
Note 2: The alternative hypothesis says "the claim is false
and what we observe happened because the claim is
false."
  o
z Test for a Population Mean
• Step 1: Write down the Null and Alternative
Hypotheses:
• Null Hypothesis
H 0 :   o
Alternative Hypothesis
H a :    o or    o or    o
Step 2
Compute the test statistic (z-score)
z
x

n
Step 3
Compute the p- value
- run a normalcdf
- For a 2 sided test you will need to double this value
The P-value is defined this way:
the probability, computed assuming the null is true, that the
test statistic would take a value as extreme or more extreme
than that actually observed is called the P-value of the test.
The smaller the P-value, the stronger the evidence against
the null hypothesis
Step 4
Compare p-value against the alpha level (  level )
-alpha level is a value that is too extreme to assume that the
sample happened just by chance
-if p-value falls below the alpha level than the null is rejected
-if p-value is above the alpha level than we fail to reject the null
• If the P-value is as small or smaller than alpha , we say that
the data are statistically significant at that level.
Radon is a colorless, odorless gas that is naturally released
by rocks and soils and may concentrate in tightly closed
houses. Because Radon is slightly radioactive, there is some
concern that it may be a health hazard. Radon detectors are
sold to home owners worried about this risk. The detectors
may be inaccurate. You placed 12 detectors in a chamber
where they were exposed to 105 picocuries per liter (pCi/l) of
Radon over three days. Here are the readings given by the
detectors:
91.9 97.8 111.4 122.3 105.4 95.0
103.8 99.6 119.3 104.8 101.7 96.6
Assume (unrealistically) that you know the standard deviation
of readings for all detectors of this type is  = 9 and that the
population of readings is approximately normal.
• (i) Give a 95% confidence interval for the mean
reading of this type of detector.
x  104.13, z  1.96,   9, n  12
*
C I  104.13  1.96(9 / 12 )
C I  (99.04, 109.23)
(ii) Is there significant evidence at the 5% level that the
mean reading differs from the true value of 105. State
the null and alternative hypotheses and conduct the
significance test.
H o :   105
H a :   105
z
104.13  105
Assume the 12 detectors are an SRS.
Already stated that normally distributed.
Ok to use normal calculations.
  0.33
9 / 12
P  value  2 norm alcdf (  100,  0.33)  0.74
Conclusion
With a P-value as high as 0.74 are sample has very good
chance of occurring if the true mean is 105. Therefore, we
do not have enough evidence to reject the null and assume
the mean reading could be around 105.
• In other words: we have no reason to think that the
average value of the readings of all such detectors is not
105 and that what we observed, namely that x = 104.13,
did not happen by chance
Suppose that in a particular geographic region, the
mean and standard deviation of scores on a reading
test are 100 points, and 12 points, respectively. Our
interest is in the scores of 55 students in a particular
school who received a mean score of 96. We can ask
whether this mean score is significantly lower than the
regional mean — that is, are the students in this school
comparable to a simple random sample of 55 students
from the region as a whole, or are their scores
surprisingly low?
z   2.47 P  value  0.0068
Homework
10.38-40