Download One Proportion z-Test

AP Stats
z-Test & Power
Homework 9
1.
One Proportion z-Test
Assume that a package of M&Ms can be considered a simple random sample of all
M&Ms.
50*0.24 = 12 ≥ 10
50*0.76 = 38 ≥ 10
Population of M&Ms is greater than 10*50 = 500
H o : The proportion of M&Ms that are blue is 24%.
H a : The proportion of M&Ms that are blue is greater than 24%.
z
p  p0
=
p0 (1  p0 )
n
11
 0.24
50
= -0.3311330893
0.24(1  0.24)
50
p-value = 0.629727957
Based upon the p-value of 0.629727957, there is not sufficient evidence to reject the null
hypothesis that the true proportion of M&Ms that are blue is 24% at any reasonable
significance level. If the Null Hypothesis were true, the probability of getting a sample
proportion of 0.22 or more extreme is 0.629727957.
2.
One Proportion z-Test
Dayton Area Drug Survey is a simple random sample.
60*0.48 = 28.8 ≥ 10
60*0.52 = 31.2 ≥ 10
Population of Dayton area seniors is greater than 10*60 = 600
H o : The proportion of Dayton area seniors that would guess that their parents feel sure
that they do not drink is 48%.
H a : The proportion of Dayton area seniors that would guess that their parents feel sure
that they do not drink is less than 48%.
z
p  p0
=
p0 (1  p0 )
n
23
 0.48
60
= -1.498753043
0.48(1  0.48)
60
p-value = 0.066968883
Based upon the p-value of 0.066968883, there is sufficient evidence to reject the null
hypothesis that the true proportion of Dayton area seniors that would guess that parents
feel sure that they do not drink is 48% at the 0.10 significance level. If the Null
Hypothesis were true, the probability of getting a sample proportion of 0.3833333 or
more extreme is 0.066968883.
3.
One Proportion z-Test
Dayton Area Drug Survey is a simple random sample.
60*0.70 = 42 ≥ 10
60*0.30 = 18 ≥ 10
Population of Dayton area seniors is greater than 10*60 = 600
H o : The proportion of Dayton area seniors that drink is 70%.
H a : The proportion of Dayton area seniors that drink is not 70%.
z
p  p0
=
p0 (1  p0 )
n
31
 0.70
60
= -3.098898934
0.70(1  0.30)
60
p-value = 0.0019425488
Based upon the p-value of 0.0019425488, there is sufficient evidence to reject the null
hypothesis that the true proportion of Dayton area seniors that drink is 70% at the 0.01
significance level. If the Null Hypothesis were true, the probability of getting a test
statistic of |-3.098898934| or more extreme is 0.0019425488.
4. It looks like it to me since 29 say that they feel sure I don’t drink or they think I
probably don’t drink and 29 also say they do not drink.
5. Type I error is a particular man is the father, but the DNA paternity test says he’s
not.
Type II error is a particular man is not the father, but the DNA paternity test says
he is.
6. Type I error is state regulators revoke the license when the repair shop really is
meeting the standards.
Type II error is the state regulators do not revoke the license when the repair shop
really is NOT meeting the standards.
7.
Test of Sample of Collection 1Test Proportion
Attribute (categorical): q53
Ho: Population proportion of 1 in q53 = 0.3
Ha: Population proportion of 1 in q53 is less
than 0.3
7 out of 40, or 0.175, are 1
z:
-1.725
P-value: 0.042
Test of Sample of Collection 1
6
Function Plot
5
4
3
2
1
0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Test Statistic
normalDensity x nullProp sampleStdDev
y = if usingNormal
sampleCount binomialProbability if round x • sampleCount
x • sampleCount
round
x • samp