STAT 200 WEEK 6 HOMEWORK SOLUTIONS 100% COMPLETE
CORRECT GUIDE
Many high school students take the AP tests in different subject areas. In 2007, of
the 144,796 students who took the biology exam, 84,199 of them were female. In
that same year, of the 211,693 students who took the Calculus AB exam 102,598 of
them were female ("AP exam scores," 2013). Estimate the difference in the
proportion of female students taking the biology exam and female students taking
the calculus AB exam using a 90% confidence level.
Given that
n1=144796,n2=211693
90% confidence interval for difference in p1-p2
=(0.5815-0.4847)+/-1.645*sqrt((0.5815*(1-0.5815)/144796)+(0.4847*(1-
0.4847)/211693))
=0.0968+/-0.0028
=(0.0941, 0.0996)
Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states
that have larger urban areas over states that are mostly rural? In the state of
Pennsylvania, a fairly urban state, there are 245 eight-year-olds diagnosed with
ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state,
there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds
evaluated ("Autism and developmental," 2008). Is there enough evidence to show
that the proportion of children diagnosed with ASD in Pennsylvania is more than
the proportion in Utah? Test at the 1% level.
Research Question:
Is there enough evidence to show that the proportion of children diagnosed
with ASD in Pennsylvania is more than the proportion in Utah?
Let p1 = the Proportion of children diagnosed with ASD in
Pennsylvania p2 = the Proportion of children diagnosed with ASD in
Utah
Null hypothesis:
H0: There is no significant difference between the proportion of children
diagnosed with ASD between Pennsylvania and Utah, p1 = p2
, Alternative hypothesis:
H1: The proportion of children diagnosed with ASD in Pennsylvania is more
than the porportion of children diagnosed with ASD in Utah, p1 > p2
Level of significance:
Critical Region:
Since the alternative hypothesis is testing on the one side (researcher is
interested if proportion in one group is more than the other) the hypothesis
is tested at one tail. Therefore, form the standard normal table for 1% level of
significance the Z-value is given as 2.33. Therefore, the null hypothesis will be
rejected if the Z-test statistic is more than 2.33
Test statistic:
Where
p1 = Proportion of children diagnosed with ASD in Pennsylvania = 245/18440
= 0.0133
p2 = Proportion of children diagnosed with ASD in Utah = 45/2123 = 0.0212
n1 = Number of children in Pennsylvania = 18440
n2 = Number of children in Utah = 2123
x1 = Number of Children diagnosed with ASD in Pennsylvania = 245
x2 = Number of Children diagnosed with ASD in Utah = 45
Therefore,
CORRECT GUIDE
Many high school students take the AP tests in different subject areas. In 2007, of
the 144,796 students who took the biology exam, 84,199 of them were female. In
that same year, of the 211,693 students who took the Calculus AB exam 102,598 of
them were female ("AP exam scores," 2013). Estimate the difference in the
proportion of female students taking the biology exam and female students taking
the calculus AB exam using a 90% confidence level.
Given that
n1=144796,n2=211693
90% confidence interval for difference in p1-p2
=(0.5815-0.4847)+/-1.645*sqrt((0.5815*(1-0.5815)/144796)+(0.4847*(1-
0.4847)/211693))
=0.0968+/-0.0028
=(0.0941, 0.0996)
Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states
that have larger urban areas over states that are mostly rural? In the state of
Pennsylvania, a fairly urban state, there are 245 eight-year-olds diagnosed with
ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state,
there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds
evaluated ("Autism and developmental," 2008). Is there enough evidence to show
that the proportion of children diagnosed with ASD in Pennsylvania is more than
the proportion in Utah? Test at the 1% level.
Research Question:
Is there enough evidence to show that the proportion of children diagnosed
with ASD in Pennsylvania is more than the proportion in Utah?
Let p1 = the Proportion of children diagnosed with ASD in
Pennsylvania p2 = the Proportion of children diagnosed with ASD in
Utah
Null hypothesis:
H0: There is no significant difference between the proportion of children
diagnosed with ASD between Pennsylvania and Utah, p1 = p2
, Alternative hypothesis:
H1: The proportion of children diagnosed with ASD in Pennsylvania is more
than the porportion of children diagnosed with ASD in Utah, p1 > p2
Level of significance:
Critical Region:
Since the alternative hypothesis is testing on the one side (researcher is
interested if proportion in one group is more than the other) the hypothesis
is tested at one tail. Therefore, form the standard normal table for 1% level of
significance the Z-value is given as 2.33. Therefore, the null hypothesis will be
rejected if the Z-test statistic is more than 2.33
Test statistic:
Where
p1 = Proportion of children diagnosed with ASD in Pennsylvania = 245/18440
= 0.0133
p2 = Proportion of children diagnosed with ASD in Utah = 45/2123 = 0.0212
n1 = Number of children in Pennsylvania = 18440
n2 = Number of children in Utah = 2123
x1 = Number of Children diagnosed with ASD in Pennsylvania = 245
x2 = Number of Children diagnosed with ASD in Utah = 45
Therefore,
