statistical survey project
TRANSCRIPT
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Statistical Survey ProjectBy: Jonathan Peñate andArnold Gonzalez
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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?
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Confidence Interval for Means
•Question 2•Question 3•Question 4•Question 5•Question 9•Question 11
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Confidence Interval for Proportions
•Question 1•Question 6•Question 7•Question 8•Question 10
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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
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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.
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Sample study links
•http://www.maletofemaleratio.com/wiki/California-CA/Baldwin_Park.htm