SOCR EduMaterials Activities Central Limit Theorem Chi square examples

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*'''Answer:'''
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<center></center>'''Answer:'''
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*a. false, the standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}</math>. Thus as the sample size increases, n increases, and as n increases, the standard deviation decreases.  
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<center></center>a. false, the standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}</math>. Thus as the sample size increases, n increases, and as n increases, the standard deviation decreases.  
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*b. True
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<center></center>b. True
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*c. False, standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}</math>
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<center></center>c. False, standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}</math>
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*d. True, the standard deviation of the total of a sample of n observations is <math>n\sqrt \sigma</math>; but the standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}.</math>Unless n is one, the standard deviation of the total of a sample of n observations exceeds the standard deviation of the sample mean.  
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<center></center>d. True, the standard deviation of the total of a sample of n observations is <math>n\sqrt \sigma</math>; but the standard deviation of the sample mean is <math>\frac{\sigma}{\sqrt n}.</math>Unless n is one, the standard deviation of the total of a sample of n observations exceeds the standard deviation of the sample mean.  
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*e. False, let's assume <math>\sigma=2</math> and <math>n=2</math>. In this case, the z-score for <math>P(\overline{X} > 4)</math> would be -2.828 while the z-score for <math>P(X>4)</math> would be -2.  <math>P(Z>-2.828) > P(Z>-2) </math>. Therefore the statement is false.  
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<center></center>e. False, let's assume <math>\sigma=2</math> and <math>n=2</math>. In this case, the z-score for <math>P(\overline{X} > 4)</math> would be -2.828 while the z-score for <math>P(X>4)</math> would be -2.  <math>P(Z>-2.828) > P(Z>-2) </math>. Therefore the statement is false.  
   
   
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*'''Answer:'''
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<center></center>'''Answer:'''
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*a. <math>P(X \ge 1000)= P(X=1000)+P(X=1001)+....+P(X=1500)= (1500 \choose 1000) \times (.7)^1000 \times (.3)^500 + (1500 \choose 1001) \times (.7)^1001 \times (.3)^499 + ...+(1500 \choose 1500) \times (.7)^1500 \times (.3)^0 = \summation (\1500 \choose X) \times (.7)^X \times (.3)^1500-X</math>
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<center></center>a. <math>P(X \ge 1000)= P(X=1000)+P(X=1001)+....+P(X=1500)= (1500 \choose 1000) \times (.7)^1000 \times (.3)^500 + (1500 \choose 1001) \times (.7)^1001 \times (.3)^499 + ...+(1500 \choose 1500) \times (.7)^1500 \times (.3)^0 = \summation (\1500 \choose X) \times (.7)^X \times (.3)^1500-X</math>
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*b. We can use the normal approximation to binomial: <math>\mu = np = 1500 \times 0.70 = 1050.</math> and <math>\sigma =  
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<center></center>b. We can use the normal approximation to binomial: <math>\mu = np = 1500 \times 0.70 = 1050.</math> and <math>\sigma =  
\sqrt npq = \sqrt1500 \times 0.7 \times 0.3= 17.748.</math>
\sqrt npq = \sqrt1500 \times 0.7 \times 0.3= 17.748.</math>
<math>P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977</math>
<math>P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977</math>
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* Below you can see a snapshot for this approximation:
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<center></center> Below you can see a snapshot for this approximation:
<center>[[Image: SOCR_Activities_CLT_Christou_example2.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_CLT_Christou_example2.jpg|600px]]</center>
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*'''Answer:'''
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<center></center>'''Answer:'''
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*a. <math>\overline{X} \sim N(8, \frac{20}{\sqrt400}). P(\overline{X} <6.50) =P(Z<-1.5)=.0667</math>
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<center></center>a. <math>\overline{X} \sim N(8, \frac{20}{\sqrt400}). P(\overline{X} <6.50) =P(Z<-1.5)=.0667</math>
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*Below you can see a snapshot for this part:
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<center></center>Below you can see a snapshot for this part:
<center>[[Image: SOCR_Activities_CLT_Christou_example3_a.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_CLT_Christou_example3_a.jpg|600px]]</center>
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*b. ??
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<center></center>b. ??
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*c. The central limit theorem states that the sample mean approaches the normal distribution as the sample size gets bigger. <center></center>Usually, if <math> n \ge 30</math>we can assume that the sample mean approaches the normal distribution. In this case <math>n=400</math>. Therefore n satisfies the requirement of a large n.   
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<center></center>c. The central limit theorem states that the sample mean approaches the normal distribution as the sample size gets bigger. Usually, if <math> n \ge 30</math> we can assume that the sample mean approaches the normal distribution. In this case <math>n=400</math>. Therefore n satisfies the requirement of a large n.   
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*d.<math>\overline{X} \sim N(8,1).</math>According to the snapshot below, the middle 80% of this distribution is (6.721,9.279). Therefore <math>w=8-6.721 =1.29</math>
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<center></center>d.<math>\overline{X} \sim N(8,1).</math>According to the snapshot below, the middle 80% of this distribution is (6.721,9.279). Therefore <math>w=8-6.721 =1.29</math>
<center>[[Image: SOCR_Activities_Normal_Christou_example3_d.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example3_d.jpg|600px]]</center>
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*e. <math>T \sim N(n\mu,\sigma\sqrt n).</math> In this case, <math>T \sim N(3200,400).</math> We know that  <math>P(T>b) =.975.</math>So now we need to find the 97.5th percentile of this distribution using SOCR. According to the SOCR snapshot below, the 97.5th percentile of this distribution is 3984. Therefore b=3984.  
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<center></center>e. <math>T \sim N(n\mu,\sigma\sqrt n).</math> In this case, <math>T \sim N(3200,400).</math> We know that  <math>P(T>b) =.975.</math>So now we need to find the 97.5th percentile of this distribution using SOCR. According to the SOCR snapshot below, the 97.5th percentile of this distribution is 3984. Therefore b=3984.  
<center>[[Image: SOCR_Activities_Normal_Christou_example3_e.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example3_e.jpg|600px]]</center>
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*'''Answer:'''
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<center></center>'''Answer:'''
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*a. <math>\overline{X} \sim N(\mu, \frac{\sigma}{\sqrt n }</math>. In this case, <math>\overline{X} \sim N(80000, 4518.48)</math>.  
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<center></center>a. <math>\overline{X} \sim N(\mu, \frac{\sigma}{\sqrt n }</math>. In this case, <math>\overline{X} \sim N(80000, 4518.48)</math>.  
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*b. We can find the answer using SOCR. The answer is 0.004032. Please see snapshot below:
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<center></center>b. We can find the answer using SOCR. The answer is 0.004032. Please see snapshot below:
<center>[[Image: SOCR_Activities_Normal_Christou_example4_.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example4_.jpg|600px]]</center>
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*c. We can find the answer right away using SOCR. Please see snapshots below:
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<center></center>c. We can find the answer right away using SOCR. Please see snapshots below:
<center>This is the distribution for <math>X</math></center>
<center>This is the distribution for <math>X</math></center>
<center>[[Image: SOCR_Activities_Normal_Christou_example4_c.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example4_c.jpg|600px]]</center>
<center>This is the distribution for <math>\overline{X}</math></center>
<center>This is the distribution for <math>\overline{X}</math></center>
<center>[[Image: SOCR_Activities_Normal_Christou_example3_c2.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example3_c2.jpg|600px]]</center>
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*The probabilities are 55.6% for  one hour vs. 86.6% for sample mean. Therefore the sample mean is more likely to be greater than 75000 hours.
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<center></center>The probabilities are 55.6% for  one hour vs. 86.6% for sample mean. Therefore the sample mean is more likely to be greater than 75000 hours.
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*'''Answer:'''
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<center></center>'''Answer:'''
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*a. According to the SOCR snapshot below, the 75th percentile is 0.006115.
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<center></center>a. According to the SOCR snapshot below, the 75th percentile is 0.006115.
<center>[[Image: SOCR_Activities_Normal_Christou_example5_aa.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example5_aa.jpg|600px]]</center>
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*b.<math>P(X>.01)=.13.</math>We can see this in the snapshot below:
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<center></center>b.<math>P(X>.01)=.13.</math>We can see this in the snapshot below:
<center>[[Image: SOCR_Activities_Normal_Christou_example5_bb.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example5_bb.jpg|600px]]</center>
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*<math>P(R=2)= (5 \choose 2) \times .13^2 \times .87^3= 0.11128. </math>
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<center></center><math>P(R=2)= (5 \choose 2) \times .13^2 \times .87^3= 0.11128. </math>
<center>[[Image: SOCR_Activities_Normal_Christou_example5_bc.jpg|600px]]</center>
<center>[[Image: SOCR_Activities_Normal_Christou_example5_bc.jpg|600px]]</center>
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*c. <center></center>  
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<center></center>c. <center></center>  
i. <math>X \sim N(.00032,.00192)</math>
i. <math>X \sim N(.00032,.00192)</math>
<center></center>
<center></center>

Revision as of 21:51, 14 May 2007

Answer:
a. false, the standard deviation of the sample mean is \frac{\sigma}{\sqrt n}. Thus as the sample size increases, n increases, and as n increases, the standard deviation decreases.
b. True
c. False, standard deviation of the sample mean is \frac{\sigma}{\sqrt n}
d. True, the standard deviation of the total of a sample of n observations is n\sqrt \sigma; but the standard deviation of the sample mean is \frac{\sigma}{\sqrt n}.Unless n is one, the standard deviation of the total of a sample of n observations exceeds the standard deviation of the sample mean.
e. False, let's assume σ = 2 and n = 2. In this case, the z-score for P(\overline{X} > 4) would be -2.828 while the z-score for P(X > 4) would be -2. P(Z > − 2.828) > P(Z > − 2). Therefore the statement is false.
Answer:
a. Failed to parse (syntax error): P(X \ge 1000)= P(X=1000)+P(X=1001)+....+P(X=1500)= (1500 \choose 1000) \times (.7)^1000 \times (.3)^500 + (1500 \choose 1001) \times (.7)^1001 \times (.3)^499 + ...+(1500 \choose 1500) \times (.7)^1500 \times (.3)^0 = \summation (\1500 \choose X) \times (.7)^X \times (.3)^1500-X
b. We can use the normal approximation to binomial: \mu = np = 1500 \times 0.70 = 1050. and \sigma = \sqrt npq = \sqrt1500 \times 0.7 \times 0.3= 17.748.

P(X \ge 1000)= P(Z> \frac{999.5-1050}{17.748}=P(Z>-2.845)=.9977

Below you can see a snapshot for this approximation:
Answer:
a. \overline{X} \sim N(8, \frac{20}{\sqrt400}). P(\overline{X} <6.50) =P(Z<-1.5)=.0667
Below you can see a snapshot for this part:
b. ??
c. The central limit theorem states that the sample mean approaches the normal distribution as the sample size gets bigger. Usually, if n \ge 30 we can assume that the sample mean approaches the normal distribution. In this case n = 400. Therefore n satisfies the requirement of a large n.
d.\overline{X} \sim N(8,1).According to the snapshot below, the middle 80% of this distribution is (6.721,9.279). Therefore w = 8 − 6.721 = 1.29
e. T \sim N(n\mu,\sigma\sqrt n). In this case, T \sim N(3200,400). We know that P(T > b) = .975.So now we need to find the 97.5th percentile of this distribution using SOCR. According to the SOCR snapshot below, the 97.5th percentile of this distribution is 3984. Therefore b=3984.
Answer:
a. \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt n }. In this case, \overline{X} \sim N(80000, 4518.48).
b. We can find the answer using SOCR. The answer is 0.004032. Please see snapshot below:
c. We can find the answer right away using SOCR. Please see snapshots below:
This is the distribution for X
This is the distribution for \overline{X}
The probabilities are 55.6% for one hour vs. 86.6% for sample mean. Therefore the sample mean is more likely to be greater than 75000 hours.
Answer:
a. According to the SOCR snapshot below, the 75th percentile is 0.006115.
b.P(X > .01) = .13.We can see this in the snapshot below:
P(R=2)= (5 \choose 2) \times .13^2 \times .87^3= 0.11128.
c.

i. X \sim N(.00032,.00192)

ii. P(\overline{X}>.005)=.0074

iii.One day's return is more likely to be greater than .007. The probabilities are 0.21 for X vs. .00022 for \overline{X}.

This is the snapshot for P(X > .007)
This is the snapshot for P(\overline{X}>.007)
Retrieved from "http://wiki.stat.ucla.edu/socr/index.php/SOCR_EduMaterials_Activities_Central_Limit_Theorem_Chi_square_examples"
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