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Math 113
Chapters 7-8 Review
#2
Identify which statement is the null hypothesis and which
is the alternative hypothesis.
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A corpse is dead
A corpse is alive
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The car has gasoline in it
The car is out of gas
#3
Identify which statement is the type I error and which is
the type II error.
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An unnecessary hysterectomy is performed
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A hysterectomy is not performed when it should be
#4
All hypothesis testing is done under the assumption that
the null hypothesis is true.
#7
A Kolmogorov-Smirnov test is performed on a sample
and the following p-values are found.
Distribution
Normal
Uniform
Exponential
Jones Fictional
p-value
0.0321
0.3215
0.1246
0.0257
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Does the sample appear to be Uniformly distributed?
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Does the sample appear to have a Jones Fictional
Distribution?
#9
Identify the test as left-tailed, right-tailed, or two-tailed
and give the decision (reject or fail to reject H0)
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If there are two critical values, it is a two-tail test.
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If the critical value is less than the mean, it is a lefttail test.
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If the critical value is greater than the mean, it is a
right-tail test.
#9 continued
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The mean of a normal or student’s t distribution is 0.
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The mean of a chi-squared distribution is its degrees
of freedom.
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The mean of an F distribution is approximately 1.
Reject the null hypothesis if the test statistic is more
extreme (in the direction of the type of test) than the
critical value – that is, if it lies in the critical region.
#9 continued again
Critical Value: t = -3.215
Test Statistic: t = -2.843
Critical Value:
Test Statistic:
df
 2 = 35.832
 2 = 41.721
= 18
#10
Harry claims that there is no agreement (null hypothesis)
between the judges of the skating competition. Kendall’s
concordance of agreement was computed and the p-value
was found to be 0.1275.
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The decision is to (reject / fail to reject) the null
hypothesis.
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There is (sufficient / insufficient) evidence to (support
/ reject) the claim that there is no agreement.
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There is (sufficient / insufficient) evidence to (support
/ reject) the claim that there is agreement.
#11
Give the decision based on the p-value.
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Reject the null hypothesis when the p-value is less
than the level of significance.
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Fail to reject the null hypothesis when the p-value is
greater than the level of significance
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P-value=0.3158, alpha = 0.3927
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P-value=0.1253, alpha = 0.1025
#28
Write the null and alternative hypothesis and identify the
test as left tailed, right tailed, or two-tailed.
1. 24% of Americans smoke
2. The average smoker began smoking before age 16
3. Professional baseball players earn more than
professional football players
4. The standard deviation on the last test was 15.
5. More women than men read Cosmopolitan.
#28 continued
1. 24% of Americans smoke
Claim : p  0.24  H 0 
H 0 : p  0.24
H1 : p  0.24  two-tail 
2. The average smoker
began smoking before
age 16
Claim:   16  H1 
H 0 :   16
H1 :   16  left tail 
#28 continued
3. Professional baseball
players earn more than
professional football
players
4. The standard deviation
on the last test was 15
Claim:  b   f  H1 
H 0 : b   f
H1 : b   f  right tail 
Claim:   15  H 0 
H 0 :   15
H1 :   15  two-tail 
#28 continued
5. More women than men
read Cosmopolitan
Claim: pw  pm  H1 
H 0 : p w  pm
H1 : pw  pm  right 
#30 - #33
Using Microsoft Excel to find critical values
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The functions and the definitions of those functions
are given on the test.
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Be sure to enter the equal sign at the beginning of the
function
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Be sure you pay careful attention to the definition.
Sometimes Excel expects a right tail probability,
sometimes a left tail probability, and sometimes a
two-tail probability. The critical value notation given
on the test is ALWAYS a RIGHT TAIL area.
#30-#33 continued
2

Find 18,0.83
=CHIINV(prob,df)
Returns the critical value from the chi-squared
distribution with df degrees of freedom and prob area in
the right tail.
Since we are looking for a right tail area (always in the
critical value notation) and Excel is expecting a right tail
area, we can just enter the function as
=CHIINV(0.83,18)
and Excel gives us the value 12.325 after we hit enter
#34
A test statistic from a sample of n=16 items is found to be
t=1.531. If this is a right tail test, find the two values that
the probability value lies between.
One Tail
Two Tail
0.025
0.05
0.05
0.10
0.10
0.20
0.25
0.50
df=15
2.131
1.753
1.341
0.691
Since the test statistic is between 1.753 and 1.341 and it
is a one-tail test, the p-value is between 0.05 and 0.10,
the one-tail probabilities associated with each of those
critical values.
0.05 < p-value < 0.10