Statistical Survey ProjectBy: Jonathan Peñate andArnold 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) = 9.87410.054-1.96(.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) = 15.33015.543+ 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.4How 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) = 5.3795.889+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) = 3.6633.990+ 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= malesp= .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= forp= .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. Census10% rule: The sample consists of less than 10% of the populationnp & nq= (.467)(92) > 10; (.499)(76899) > 10 np &nq= (.537)(92) > 10; (.501)(76899) > 10
3. We will conduct a 2-proportion z- test
4. 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 populationnp & nq= (.473)(91) > 10; (.527)(91) > 10np & 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 sample10% rule: We have less than 10% of the populationNearly 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
Observed Yes No Omitted Total
Freshman 27 2 2 31
Sophomore 24 6 0 30
Junior 27 2 0 29
Senior 2 0 1 3
Total 80 10 3 93
Observed Yes No Omitted
Freshman 26.667 3.333 1
Sophomore 25.806 3.226 .9677
Junior 24.946 3.1183 .935
Senior 2.581 .323 .097
Chi Squared Test for Homogeneity Q.6 Continued
1. Ho: Responses are independent of grade levelHa: Responses are not independent of grade level
2. Randomness: We conducted a random sample10% 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