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Statistical Survey Project By: Jonathan Peñate  and Arnold Gonzalez
Our Survey Questions 1. What is your gender? 2. What grade are you in? 3. What is your current age? 4. How many people live in your household (including yourself?)  5. How many pets do you own? 6. Do you like your parents? 7. Are you for or against the legalization of marijuana?  8. Are you in a school sport? 9. If yes, how many? 10. Are there any video game consoles in your house?  11. How many TV’s are in your household?
Confidence Interval for Means Question 2 Question 3 Question 4 Question 5 Question 9 Question 11
Confidence Interval for Proportions Question 1 Question 6 Question 7 Question 8 Question 10
Confidence Interval for Mean Q.2 What grade are you in? X= 10.054, St. Dev= .882, n=92, df=91,  z*=1.96 10.054-1.96(.882/sqt 92) = 9.874 10.054-1.96(.882/sqt 92) = 10.234 Confidence Interval: (9.874, 10.234) We are 95% confident that the true mean grade of the students is between 9.874 and 10.234
Confidence Interval for Mean Q.3 What is your current age? X= 15.543,  St. Dev= 1.042, n=92, df=91, z*=1.96 15.543- 1.96(1.042/sqt 92) = 15.330 15.543+ 1.96(1.042/.sqt 92) = 15.756 Confidence Interval: (15.330, 15.756) We are 95% confident that the true mean age of the students is between 15.330 and 15.756
Confidence Interval Q.4 How many people live in your household? X= 5.889,  St. Dev= 2.470,  n=90, df= 89, z*=1.96 5.889- 1.96(2.470/sqt 90) = 5.379 5.889+1.96(2.470/sqt 90) =6.399 Confidence Interval: (5.379, 6.399) We are 95% confident that the true mean of people living in the students’ households is between 5.379 and 6.399
Confidence Interval for Mean Q.5 How many pets do you own? X= 1.441, St. Dev= 1.515, n= 92, df= 91, z*= 1.96 1.441- 1.96(1.441/stq 92) = 1.314  1.441+ 1.96(1.441/stq 92)= 1.751 Confidence Interval: (1.314, 1.751) We are 95% confident that the true mean of pets owned by the students is between 1.314 and 1.751
Confidence Interval for Mean Q.9 How many school sports are you in? X= .359, St. Dev= .585, n=92, df= 91, z*1.96 .359- 1.96(.585/sqt 92) = .239 .359+ 1.96(.585/sqt 92)= .479 Confidence Interval: (.239, .479) We are 95% that the true mean of students who are in a school sport is between .239 and .479
Confidence Interval for Mean Q.11 How many TV’s are in your household? X= 3.990, St. Dev= 1.600, n= 92, df= 91, z*= 1.96 3.990- 1.96(1.600/stq 92) = 3.663 3.990+ 1.96(1.600/stq 92) = 4.317 Confidence Interval: (3.663, 4.317) We are 95% confident that the true mean of TV’s in students household is between 3.663 and 4.317
Confidence Interval for Proportion Q.1 What is your gender?  p= males p= .467, q= .532,  z*=1.96, n=92 .467- 1.96*sqt[(.467)(.532)/92]= .365 .467+ 1.96*sqt[(.467)(.532)/92] = .569 Confidence Interval= (.365, .569) We are 95% that the proportion of male students is between .365 and .569
Confidence Interval for Proportion Q.6 Do you like you parents?  p= .887, q= .123, z*= 1.96, n=89 .887- 1.96*sqt[(.822)(.178)/89)] = .822 .887+ 1.96*sqt[(.822)(.178)/89)] = .953 Confidence Interval: (.822, .953) We are 95% confident that the true proportion of students who like their parents is between .822 and .953
Confidence Interval for Proportion Q.7 Are you for or against marijuana? p= for p= .473, q= .527, z*= 1.96, n= 91 .473- 1.96*sqt[(.473)(.527)/(89)]= .369 .473+ 1.96*sqt[(.473)(.527)/(89)]= .575 Confidence Interval: (.369, .575) We are 95% confident that the true proportion of students who are for marijuana are between .369 and .575
Confidence Interval for Proportion Q.8 Are you in a school sport? p= .696, q= .304, z*= 1.96, n= 92 .696- 1.96*sqt[(.696)(.304)/92]= .376 .696+ 1.96*sqt[(.696)(.304)/92]= .583 Confidence Interval: (.376, .583) We are 95% confident that the true proportion of students who are in a school sport is between .376 and .583
Confidence Interval from Proportion Q.10  Are there any video game consoles in your house? p= .867, q= .133, z*= 1.96, n= 90 .867- 1.96* sqt[(.867)(.133)/90]= .063 .867+ 1.96* sqt[(.867)(.133)/90]= .203 Confidence Interval: (.063, .203) We are 95% confident that the true proportion of students who own have a video game console in their house is between .063 and .203
Hypothesis test for Q.1 vs Larger Study 1. Ho: p1 = p2; Ha: p1 ≠ p2 2. Randomness: The study is from the U.S. Census 10% rule: The sample consists of less than 10% of the population np & nq= (.467)(92) > 10; (.499)(76899) > 10 np &nq= (.537)(92) > 10; (.501)(76899) > 10 3. We will conduct a 2-proportion z- test Do the math… z= -.6128; p= .54 5. With such a high probability we fail to reject the null hypothesis. Therefore, there  is not enough evidence to say that there is a difference between the two proportions.
Hypothesis test for Q.7 vs Larger Study  1. Ho: p1 = p2; Ha: p1 ≠ p2 2. Randomness: Randomness is not stated. 10% rule: There is less than 10% of the total population np & nq= (.473)(91) > 10; (.527)(91) > 10 np & nq= (.450)(828) > 10; (.500)(828) > 10 3. We will conduct a 2 proportion z- test 4. Do the math… z= -.4975; p= .618 5. With such a high probability, we fail to reject the Ho. There is not enough evidence to say that there is a difference between the two proportions.
Hypothesis on affirmative responses (Males vs Females) for Q.6 1. H0:  μ of males = μ of females; Ha: μ of males   ≠ μ of females. 2. Randomness: This was  a random sample 10% rule: We have less than 10% of the population Nearly normal: We can assume  the data is normally distributed. 3. We will conduct a 2 sample  t- test 4. Do the math… t= .083; df= 3.88, p= .937 5. With such a high probability, we fail to reject the null hypothesis. There is not enough evidence to say  that there is a difference in the two sample means of affirmative answers.
Chi Squared Test for Homogeneity Q.6
Chi Squared Test for Homogeneity Q.6 Continued 1. Ho: Responses are independent of grade level Ha: Responses are not independent of grade level 2. Randomness: We conducted a random sample 10% condition: We have less than 10% of the population 3. We will conduct a X^2 test for homegeneity 4. Do the math… Chi^2 = 15.41; p= .22 5. Based on such a high probability, we cannot  reject the null  hypothesis. There is not enough evidence to say responses are not independent of grade level.
Sample study links http://www.maletofemaleratio.com/wiki/California-CA/Baldwin_Park.htm

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Statistical survey project

  • 1. Statistical Survey Project By: Jonathan Peñate and Arnold Gonzalez
  • 2. Our Survey Questions 1. What is your gender? 2. What grade are you in? 3. What is your current age? 4. How many people live in your household (including yourself?) 5. How many pets do you own? 6. Do you like your parents? 7. Are you for or against the legalization of marijuana? 8. Are you in a school sport? 9. If yes, how many? 10. Are there any video game consoles in your house? 11. How many TV’s are in your household?
  • 3. Confidence Interval for Means Question 2 Question 3 Question 4 Question 5 Question 9 Question 11
  • 4. Confidence Interval for Proportions Question 1 Question 6 Question 7 Question 8 Question 10
  • 5. Confidence Interval for Mean Q.2 What grade are you in? X= 10.054, St. Dev= .882, n=92, df=91, z*=1.96 10.054-1.96(.882/sqt 92) = 9.874 10.054-1.96(.882/sqt 92) = 10.234 Confidence Interval: (9.874, 10.234) We are 95% confident that the true mean grade of the students is between 9.874 and 10.234
  • 6. Confidence Interval for Mean Q.3 What is your current age? X= 15.543, St. Dev= 1.042, n=92, df=91, z*=1.96 15.543- 1.96(1.042/sqt 92) = 15.330 15.543+ 1.96(1.042/.sqt 92) = 15.756 Confidence Interval: (15.330, 15.756) We are 95% confident that the true mean age of the students is between 15.330 and 15.756
  • 7. Confidence Interval Q.4 How many people live in your household? X= 5.889, St. Dev= 2.470, n=90, df= 89, z*=1.96 5.889- 1.96(2.470/sqt 90) = 5.379 5.889+1.96(2.470/sqt 90) =6.399 Confidence Interval: (5.379, 6.399) We are 95% confident that the true mean of people living in the students’ households is between 5.379 and 6.399
  • 8. Confidence Interval for Mean Q.5 How many pets do you own? X= 1.441, St. Dev= 1.515, n= 92, df= 91, z*= 1.96 1.441- 1.96(1.441/stq 92) = 1.314 1.441+ 1.96(1.441/stq 92)= 1.751 Confidence Interval: (1.314, 1.751) We are 95% confident that the true mean of pets owned by the students is between 1.314 and 1.751
  • 9. Confidence Interval for Mean Q.9 How many school sports are you in? X= .359, St. Dev= .585, n=92, df= 91, z*1.96 .359- 1.96(.585/sqt 92) = .239 .359+ 1.96(.585/sqt 92)= .479 Confidence Interval: (.239, .479) We are 95% that the true mean of students who are in a school sport is between .239 and .479
  • 10. Confidence Interval for Mean Q.11 How many TV’s are in your household? X= 3.990, St. Dev= 1.600, n= 92, df= 91, z*= 1.96 3.990- 1.96(1.600/stq 92) = 3.663 3.990+ 1.96(1.600/stq 92) = 4.317 Confidence Interval: (3.663, 4.317) We are 95% confident that the true mean of TV’s in students household is between 3.663 and 4.317
  • 11. Confidence Interval for Proportion Q.1 What is your gender? p= males p= .467, q= .532, z*=1.96, n=92 .467- 1.96*sqt[(.467)(.532)/92]= .365 .467+ 1.96*sqt[(.467)(.532)/92] = .569 Confidence Interval= (.365, .569) We are 95% that the proportion of male students is between .365 and .569
  • 12. Confidence Interval for Proportion Q.6 Do you like you parents? p= .887, q= .123, z*= 1.96, n=89 .887- 1.96*sqt[(.822)(.178)/89)] = .822 .887+ 1.96*sqt[(.822)(.178)/89)] = .953 Confidence Interval: (.822, .953) We are 95% confident that the true proportion of students who like their parents is between .822 and .953
  • 13. Confidence Interval for Proportion Q.7 Are you for or against marijuana? p= for p= .473, q= .527, z*= 1.96, n= 91 .473- 1.96*sqt[(.473)(.527)/(89)]= .369 .473+ 1.96*sqt[(.473)(.527)/(89)]= .575 Confidence Interval: (.369, .575) We are 95% confident that the true proportion of students who are for marijuana are between .369 and .575
  • 14. Confidence Interval for Proportion Q.8 Are you in a school sport? p= .696, q= .304, z*= 1.96, n= 92 .696- 1.96*sqt[(.696)(.304)/92]= .376 .696+ 1.96*sqt[(.696)(.304)/92]= .583 Confidence Interval: (.376, .583) We are 95% confident that the true proportion of students who are in a school sport is between .376 and .583
  • 15. Confidence Interval from Proportion Q.10 Are there any video game consoles in your house? p= .867, q= .133, z*= 1.96, n= 90 .867- 1.96* sqt[(.867)(.133)/90]= .063 .867+ 1.96* sqt[(.867)(.133)/90]= .203 Confidence Interval: (.063, .203) We are 95% confident that the true proportion of students who own have a video game console in their house is between .063 and .203
  • 16. Hypothesis test for Q.1 vs Larger Study 1. Ho: p1 = p2; Ha: p1 ≠ p2 2. Randomness: The study is from the U.S. Census 10% rule: The sample consists of less than 10% of the population np & nq= (.467)(92) > 10; (.499)(76899) > 10 np &nq= (.537)(92) > 10; (.501)(76899) > 10 3. We will conduct a 2-proportion z- test Do the math… z= -.6128; p= .54 5. With such a high probability we fail to reject the null hypothesis. Therefore, there is not enough evidence to say that there is a difference between the two proportions.
  • 17. Hypothesis test for Q.7 vs Larger Study 1. Ho: p1 = p2; Ha: p1 ≠ p2 2. Randomness: Randomness is not stated. 10% rule: There is less than 10% of the total population np & nq= (.473)(91) > 10; (.527)(91) > 10 np & nq= (.450)(828) > 10; (.500)(828) > 10 3. We will conduct a 2 proportion z- test 4. Do the math… z= -.4975; p= .618 5. With such a high probability, we fail to reject the Ho. There is not enough evidence to say that there is a difference between the two proportions.
  • 18. Hypothesis on affirmative responses (Males vs Females) for Q.6 1. H0: μ of males = μ of females; Ha: μ of males ≠ μ of females. 2. Randomness: This was a random sample 10% rule: We have less than 10% of the population Nearly normal: We can assume the data is normally distributed. 3. We will conduct a 2 sample t- test 4. Do the math… t= .083; df= 3.88, p= .937 5. With such a high probability, we fail to reject the null hypothesis. There is not enough evidence to say that there is a difference in the two sample means of affirmative answers.
  • 19. Chi Squared Test for Homogeneity Q.6
  • 20. Chi Squared Test for Homogeneity Q.6 Continued 1. Ho: Responses are independent of grade level Ha: Responses are not independent of grade level 2. Randomness: We conducted a random sample 10% condition: We have less than 10% of the population 3. We will conduct a X^2 test for homegeneity 4. Do the math… Chi^2 = 15.41; p= .22 5. Based on such a high probability, we cannot reject the null hypothesis. There is not enough evidence to say responses are not independent of grade level.
  • 21. Sample study links http://www.maletofemaleratio.com/wiki/California-CA/Baldwin_Park.htm