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STAT 200 Week 6 Homework Problems9.1.2
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.
Population 1 (Biology)
𝒏𝟏 = 𝟏𝟒𝟒𝟕𝟗𝟔
𝒙𝟏 = 𝟖𝟒𝟏𝟗𝟗
Population 2 (Calculus AB)
𝒏𝟐 = 𝟐𝟏𝟏𝟔𝟗𝟑
𝒙𝟐 = 𝟏𝟎𝟐𝟓𝟗𝟖
̂𝟏 − 𝒑
𝒑 ̂𝟐 − (𝒑𝟏 − 𝒑𝟐 )
𝒛=
𝒑 (𝟏 − 𝒑𝟏 ) 𝒑𝟐 (𝟏 − 𝒑𝟐 )
√ 𝟏 +
𝒏𝟏 𝒏𝟐
Sample 1 proportion:
𝒙𝟏 𝟖𝟏𝟒𝟗𝟗
̂𝟏 =
𝒑 = = 𝟎. 𝟓𝟖𝟏𝟓
𝒏𝟏 𝟏𝟒𝟒𝟕𝟗𝟔
Sample 2 proportion:
𝒙𝟐 𝟏𝟎𝟐𝟓𝟗𝟖
̂𝟐 =
𝒑 = = 𝟎. 𝟒𝟖𝟒𝟕
𝒏𝟐 𝟐𝟏𝟏𝟔𝟗𝟑
Point Estimation:
̂𝟏 − 𝒑
𝒑𝟏 − 𝒑𝟐 = 𝒑 ̂𝟐 = 𝟎. 𝟓𝟖𝟏𝟓 − 𝟎. 𝟒𝟖𝟒𝟕 = 𝟎. 𝟎𝟗𝟔𝟖
𝟏 − 𝜶 = 𝟎. 𝟗𝟎
𝜶 = 𝟎. 𝟏𝟎
𝜶
= 𝟎. 𝟎𝟓
𝟐
From the table, the closest values to 0.9500 are 1.6 and 0.05.
𝒛𝜶⁄𝟐 = 𝟏. 𝟔 + 𝟎. 𝟎𝟓 = 𝟏. 𝟔𝟓
Calculating the confidence interval:
, ̂(𝟏−𝒑
𝒑 𝟏 ̂)
𝟏 ̂(𝟏−𝒑
𝒑 ̂)
̂𝟏 − 𝒑
CI = 𝒑 ̂𝟐 ± 𝒛𝜶⁄𝟐 ∗ √
𝒏𝟏
+ 𝟐𝒏 𝟐
𝟐
𝟎.𝟓𝟖𝟏𝟓(𝟏−𝟎.𝟓𝟖𝟏𝟓) 𝟎.𝟒𝟖𝟒𝟕(𝟏−𝟎.𝟒𝟖𝟒𝟕)
CI = 𝟎. 𝟓𝟖𝟏𝟓 − 𝟎. 𝟒𝟖𝟒𝟕 ± 𝟏. 𝟔𝟓 ∗ √ 𝟏𝟒𝟒𝟕𝟗𝟔
+ 𝟐𝟏𝟏𝟔𝟗𝟑
𝟎.𝟓𝟖𝟏𝟓(𝟎.𝟒𝟏𝟖𝟓) 𝟎.𝟒𝟖𝟒𝟕(𝟎.𝟓𝟏𝟓𝟑)
CI = 𝟎. 𝟎𝟗𝟔𝟖 ± 𝟏. 𝟔𝟓 ∗ √ 𝟏𝟒𝟒𝟕𝟗𝟔
+ 𝟐𝟏𝟏𝟔𝟗𝟑
𝟎.𝟐𝟒𝟑𝟑𝟓𝟕𝟕𝟓 𝟎.𝟐𝟒𝟗𝟕𝟔𝟓𝟗𝟏
CI = 𝟎. 𝟎𝟗𝟔𝟖 ± 𝟏. 𝟔𝟓 ∗ √ 𝟏𝟒𝟒𝟕𝟗𝟔
+ 𝟐𝟏𝟏𝟔𝟗𝟑
CI = 𝟎. 𝟎𝟗𝟔𝟖 ± 𝟏. 𝟔𝟓 ∗ √𝟏. 𝟔𝟖𝟎𝟔𝟗 ∗ 𝟏𝟎−𝟔 + 𝟏. 𝟏𝟕𝟗𝟖𝟓 ∗ 𝟏𝟎−𝟔
CI = 𝟎. 𝟎𝟗𝟔𝟖 ± 𝟏. 𝟔𝟓 ∗ √𝟐. 𝟖𝟔𝟎𝟓𝟒 ∗ 𝟏𝟎−𝟔
CI = 𝟎. 𝟎𝟗𝟔𝟖 ± 𝟏. 𝟔𝟓 ∗ 𝟎. 𝟎𝟎𝟏𝟔𝟗𝟏𝟑𝟏𝟒
CI = 𝟎. 𝟎𝟗𝟔𝟖 ± 𝟎. 𝟎𝟑
CI = (0.094, 0.100)
9.1.5
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.
Sample Proportions:
𝟐𝟒𝟓
̂𝑷 =
𝒑 = 𝟎. 𝟎𝟏𝟑𝟑
𝟏𝟖𝟒𝟒𝟎
𝟒𝟓
𝒑̂𝑼 = = 𝟎. 𝟎𝟐𝟏𝟐𝟎
𝟐𝟏𝟐𝟑
Testing if there is enough evidence to show that the proportion of children diagnosed in Pennsylvania
is more than the proportion of children diagnosed in Utah, at the 1% level:
Null hypothesis: 𝒑𝑷 ≤ 𝒑𝑼
Alternate hypothesis: 𝒑𝑷 > 𝒑𝑼
𝟐𝟒𝟓+𝟒𝟓
̅=
Pooled proportion: 𝒑 = 𝟎. 𝟎𝟏𝟒𝟏
𝟏𝟖𝟒𝟒𝟎+𝟐𝟏𝟐𝟑
Test statistic:
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