AP Statistics Curriculum 2007 StudentsT - Revision history

Tdlee: Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

2020-03-03T19:28:21Z

Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

Dinov: /* Approach III (Approximate) */

2014-08-04T20:10:02Z

‎Approach III (Approximate)

Dinov: /* Approach III (Approximate) */

2014-08-04T20:09:45Z

‎Approach III (Approximate)

Dinov: /* Approach II (Approximate) */

2014-08-04T20:09:16Z

‎Approach II (Approximate)

Dinov: /* Approach II (Approximate) */

2014-08-04T20:08:36Z

‎Approach II (Approximate)

Dinov: /* Approach I (exact) */

2014-08-04T20:08:08Z

‎Approach I (exact)

Dinov: /* Approach I (exact) */

2014-08-04T20:07:50Z

‎Approach I (exact)

IvoDinov: /* Approach III (Approximate) */

2011-03-11T00:24:15Z

‎Approach III (Approximate)

IvoDinov: /* Approach I (exact) */

2011-02-16T16:48:11Z

‎Approach I (exact)

Jenny at 20:31, 28 June 2010

2010-06-28T20:31:56Z

← Older revision		Revision as of 19:28, 3 March 2020
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	* SOCR Home page: http://www.socr.ucla.edu		* SOCR Home page: http://www.socr.ucla.edu

−	{{translate\|pageName=http://wiki.~~stat~~.~~ucla~~.edu~~/socr~~/index.php?title=AP_Statistics_Curriculum_2007_StudentsT}}	+	"{{translate\|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_StudentsT}}

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 ====Approach III (Approximate)====
-: <math>X \sim B(n=12, p=0.44)</math>. The sample proportion is <math>\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433].</math> Thus, <math>P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2),</math> where <math>p_1=0.5-1/24</math> and <math>p_2=0.5+1/24.</math> Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits (<math>p_1</math> and <math>p_2</math>), using the <math>Z = (p-0.44)/0.1433</math> transformation, to get
+: <math>X \sim B(n=12, p=0.44).</math> The sample proportion is <math>\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433].</math> Thus, <math>P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2),</math> where <math>p_1=0.5-1/24</math> and <math>p_2=0.5+1/24.</math> Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits (<math>p_1</math> and <math>p_2</math>), using the <math>Z = (p-0.44)/0.1433</math> transformation, to get
 :<math>P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.</math>

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 ====Approach III (Approximate)====
-: <math>X \sim B(n=12, p=0.44)</math>. The sample proportion is <math>\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433]</math>. Thus, <math>P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2)</math>, where <math>p_1=0.5-1/24</math> and <math>p_2=0.5+1/24</math>. Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits (<math>p_1</math> and <math>p_2</math>), using the <math>Z = (p-0.44)/0.1433</math> transformation, to get
+: <math>X \sim B(n=12, p=0.44)</math>. The sample proportion is <math>\hat{p} = X/n \approx N [m = p = 0.44; (p(1-p)/n)1/2=0.1433].</math> Thus, <math>P(X=6) = P(\hat{p}=0.5) \approx P(p_1 \leq \hat{p} \leq p_2),</math> where <math>p_1=0.5-1/24</math> and <math>p_2=0.5+1/24.</math> Note that approach II is very similar to approach III, however, the former uses the total sum of successes (X), whereas the latter employs the proportion (X/n). This is why the left-right additive term of 0.5 in approach II becomes a 0.5*(1/12) = 1/24 in the III approximation. Finally, standardize each of the 2 limits (<math>p_1</math> and <math>p_2</math>), using the <math>Z = (p-0.44)/0.1433</math> transformation, to get
 :<math>P(X=6) \approx P(p_1 \leq \hat{p} \leq p_2) = P(Z_1 \leq Z \leq Z_2) = 0.211.</math>
 <hr>

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 ====Approach II (Approximate)====
 : <math>X \sim B(n=12, p=0.44).</math>
-: <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7]</math>. <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X-5.28} \over {1.7}}</math> and <math>X_1=5.5,</math> <math>X_2=6.5.</math> So, <math>P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.</math>
+: <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7].</math> <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X-5.28} \over {1.7}}</math> and <math>X_1=5.5,</math> <math>X_2=6.5.</math> So, <math>P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.</math>
 ====Approach III (Approximate)====

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 ====Approach II (Approximate)====
-: <math>X \sim B(n=12, p=0.44)</math>.
+: <math>X \sim B(n=12, p=0.44).</math>
-: <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7]</math>. <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X-5.28} \over {1.7}}</math> and <math>X_1=5.5</math>, <math>X_2=6.5</math>. So, <math>P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.</math>
+: <math>X (approx.) \sim N [\mu = n p = 5.28; \sigma=\sqrt{(np(1-p))}=1.7]</math>. <math>P(X=6) \approx P(Z_1\leq Z \leq Z_2)</math>, where <math>Z = {{X-5.28} \over {1.7}}</math> and <math>X_1=5.5,</math> <math>X_2=6.5.</math> So, <math>P(X=6)\approx P(Z_1 \leq Z \leq Z_2)=0.211.</math>
 ====Approach III (Approximate)====

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 ====Approach I (exact)====
 : <math>P(X=6)=?</math> where <math>X\sim B(12, 0.44)</math>
-: <math>P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068</math>, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator].
+: <math>P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068,</math> using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator].
 ====Approach II (Approximate)====

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 * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types. What is the probability that exactly 6 subjects have type O blood type in the sample?
 ====Approach I (exact)====
-: <math>P(X=6)=</math>? Where <math>X\sim B(12, 0.44)</math>
+: <math>P(X=6)=</math>? Where <math>X\sim B(12, 0.44)</math>
 : <math>P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068</math>, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator].

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 ==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - Student's T Distribution==
-Very frequently in practice, we do not know the population variance, therefore we need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting <math>Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}</math>.
+Very frequently in practice, we do not know the population variance. Therefore we need to estimate it using the sample-variance. This requires us to introduce the T-distribution, which is a one-parameter distribution connecting <math>Cauchy=T_{(df=1)} \longrightarrow T_{(df)}\longrightarrow N(0,1)=T_{(df=\infty)}</math>.
 [[Image:SOCR_Distribution_StudentT_density.JPG|200px|thumbnail|right]]
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 * Suppose the proportion of blood type O in the population is 0.44. If we take a random sample of 12 subjects and make a note of their blood types. What is the probability that exactly 6 subjects have type O blood type in the sample?
-====Approach I (exact!)====
+====Approach I (exact)====
 : <math>P(X=6)=</math>? Where <math>X\sim B(12, 0.44)</math>
 : <math>P(X=6)={12\choose 6}p^6(1-p)^{6}=\frac{12!}{6!(6)!}0.44^6 0.56^6=0.2068</math>, using SOCR Binomial [http://socr.ucla.edu/htmls/SOCR_Distributions.html interactive GUI] or [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm calculator].