The document analyzes statistics from a survey administered to Baldwin Park High School students. It provides confidence intervals for survey questions asking about relationships, beliefs, and preferences. Hypothesis tests are conducted to compare survey results to larger studies and compare responses between genders. Chi-square tests determine whether certain opinions vary with age. While some opinions differed significantly between genders or with prior studies, age was found to not influence most beliefs and behaviors examined.

1. Analyses of Survey’s Statistics Griselda Escobedo Jessica DeLuna Avigail Carrillo Mr. Eastvedt, Period 4

2. Survey Questions Administered to Baldwin Park High School Students: 1. Male/Female 2. Age 3. Are you single? Y/N 4. If your partner cheated on you would you react violently? Y/N 5. Do you like Justin Bieber? Y/N 6. Do you believe in God? Y/N 7. Have you ever cut class? Y/N 8. How many years older than you would you date someone? (#) 9. How many children do you want? (#) 10. On a scale of 1-10, how important is the color of a car to you? (10 being the highest)

3. Confidence Intervals Equation for numerical questions: Sample mean ± 1.96 (σ/√n) One can be 95% confident that the population mean is contained within that interval. Equation for opinion questions (proportions): p hat = proportion of “Yes” answers q hat= 1 – p hat = proportion of “No” answers p hat ± 1.96√[(p hat∙q hat)/n] One can be 95% confident that the interval will contain the population proportion.

4. Confidence intervals for the means of numerical questions: Q8: 5.339 ± 1.230 We are 95% confident that the true mean of accepted age difference is between 4.109 and 6.569. Q9: 5.273 ± 2.741 We are 95% confident that the true mean of children wanted is between 2.532 and 8.014. Q10: 6.567 ± 0.577 We are 95% confident that the true mean of the scaled importance of car color is between 5.990 and 7.144.

5. Confidence intervals for opinion questions (Y/N proportion): Q3: 0.618 ± 0.101 We are 95% confident that the population proportion of single individuals is within the interval (0.517, 0.719). Q4: 0.146 ± 0.073 We are 95% confident that the population proportion of violent reactions to infidelity is within the interval (0.073, 0.219). Q5: 0.178 ± 0.079 We are 95% confident that the population proportion of Justin Bieber fans is within the interval (0.099, 0.257)

6. Proportion confidence intervals continued.. Q6: 0.862 ± 0.072 We can be 95% confident that the population proportion is within the interval (0.790, 0.934). Q7: 0.544 ± 0.103 We can be 95% confident that the population proportion is within the interval (0.441, 0.647).

7. Question 6: Hypothesis Test on Larger Study “The survey finds that the number of people who say they are unaffiliated with any particular faith today (16.1%)..” http://religions.pewforum.org/reports The results of our survey administered to Baldwin Park High School show that 13 out of 87 (≈15%) of the students do not believe in God. 1. H0: p=.161 Ha: p≠.161 n=87 2. Assumptions/Conditions: Randomness: The sample was randomly selected. 10% condition: The sample is less than 10% of the population. We may assume the data is evenly distributed. 3. We will conduct a 1-proportion z-test 4. P hat= 0.150 z=(.150-.161)/√[(.161)(.839)/87]= -0.279 P-value is ≈ 0.3897 5. Since the p-value is significantly greater than 0.05, we may not reject the null hypothesis. There is not enough evidence to suggest that there is a significant difference between the proportion of atheists.

8. Question 7: Hypothesis Test on Larger Study “The high dropout rate may also be related to the finding that half of the respondents said they have skipped school..” http://newsinfo.iu.edu/web/page/normal/4948.html The results of our survey reveal that 40 out of 89 students (≈50%) have cut class before. 1. H0: p=.50 Ha: p≠.50 n=89 2. Assumptions/Conditions: Randomness: The sample was randomly selected. 10% condition: The sample is less than 10% of the population. We may assume the data is evenly distributed. 3. We will conduct a 1-proportion z-test 4. P hat= 0.50 z=(.50-.50)/√[(.50(.50/89]= 0 P-value is .5000 With such a high p-value, we cannot reject the null hypothesis. Therefore we may conclude that the proportions of students that have skipped school are equal.

9. Comparing the means of affirmative responses for males vs. females Question 8: How many years older than you would you date someone? 1. H0: MeanM= MeanF ; Ha: MeanM≠ MeanF; Mean= mean yrs. difference 2. -Randomness: Survey sample was acquired randomly. -Independence: The age of one student does not affect another’s. -10% condition: The 87 students are less than 10% of the population. -Nearly normal: The distributions are nearly normal. 3. We will conduct a 2 sample T-test. 4. nM=46 nF=41 XM=3.707 XF=7.171 sM=4.786 sF=6.524 df=85 t= (XM-XF)/[√(sM2/nM)+(sF2/nF)] = -2.795 p= 0.006414 5. Since the p-value is less than 0.05, we can reject the null hypothesis; there is enough evidence to suggest that mean years-difference accepted by males is different from that of females.

10. Comparing the means of affirmative responses for males vs. females Question 9: How many children do you want? H0: MeanM= MeanF; Ha: MeanM≠ MeanF; Mean= mean children wanted 2. -Randomness: Survey sample was acquired randomly. -Independence: The age of one student does not affect another’s. -10% condition: The 88 students are less than 10% of the population. -Nearly normal: The distributions are nearly normal. We will conduct a 2 sample T-test. nM=47 nF=41 XM=3.957 XF=6.780 sM=6.032 sF=18.231 df=86 t= (XM-XF)/[√(sM2/nM)+(sF2/nF)] = -0.947 p=0.346292 5. Since the p-value is less than 0.05, we may reject the null hypothesis that the mean children wanted by males is equal to that of females; there is enough evidence to prove that the difference is significant.

11. Comparing the means of affirmative responses for males vs. females Question 10: On a scale of 1-10, how important is the color of a car to you? (10 being the highest) H0: MeanM= MeanF ; Ha: MeanM≠ MeanF; Mean= mean scale 2. -Randomness: Survey sample was acquired randomly. -Independence: The age of one student does not affect another’s. -10% condition: The 89 students are less than 10% of the population. -Nearly normal: The distributions are nearly normal. 3. A 2 sample T-test will be conducted. 4. nM=48 nF=41 XM=7.115 XF=5.927 sM=2.610 sF=2.893 df=87 t= (XM-XF)/[√(sM2/nM)+(sF2/nF)] = 2.020 p=0.046460 5. There is enough evidence against the null hypothesis since the p-value is less than 0.05. We may conclude that the importance of a car’s color is different among males vs. females and reject the null hypothesis.

12. Chi2 Tests to determine whether grade levels have different opinions. Equation: X2= ∑ [ (O – E )2 / E ] where O is the observed value, and E is the expected value.

13. Chi2 Test Question 4: If your partner cheated on you would you react violently? H0: Reacting violently is independent of age. Ha: Reacting violently is not independent of age. Observed: Expected: X2=1.075 P=0.898 df=4 Since the p-value is greater than 0.05, we do not reject the null hypothesis. We may conclude that reaction violently is independent of age.

14. Chi2 Test Question 5: Do you like Justin Bieber? H0: Having a positive regard of Justin Bieber is independent of age. Ha: Having a positive regard of Justin Bieber is not independent of age. Observed: Expected: X2 =7.787 P=0.010 df=4 At the 0.05 significance level we may reject the null hypothesis and conclude that liking Justin Bieber is dependent on age.

15. Chi2 Test Question 6: Do you believe in God? H0: Believing in God is independent of age. Ha: Believing in God is not independent of age. Observed: Expected: X2=5.208 P=0.267 df=4 With such a high p-value, at the 0.05 level of significance the null hypothesis may not be rejected. Therefore we conclude that the belief in God is independent of age.

16. Chi2 Test Question 7: Have you ever cut class? H0: Having ever cut class is independent of age. Ha: Having ever cut class is not independent of age. Observed: Expected: X2=5.165 P=0.271 df=4 Since the p-value is greater than 0.05, we may not reject the null hypothesis and conclude that cutting class is independent of age.