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1. Case Study: Prevalence of Mild Anemia in School Children in a Rural Community in Peru Introduction: Mild anemia is a public health problem in Peru, especially in rural areas where access to adequate nutrition may be limited. This study seeks to investigate the prevalence of mild anemia in schoolchildren in a rural community in Peru and its possible association with access to a balanced diet. Problem: What is the prevalence of mild anemia in schoolchildren in a rural community in Peru and how is it related to access to a balanced diet? / Determine whether there is a significant association between dichotomous variables. Anemia (Present/absent). Balanced diet (Yes / No). Data were collected from 95 schoolchildren in a rural community in the country, recording their anemia status (present or absent) and their access to a balanced diet (yes or no), as assessed by parents or guardians. The data are presented in the following table: Table 1: Type of diet and anemia in children in a rural community, Peru 2022. Balanced diet Anemia Present Absent No 32 12 Yes 9 42 With a significance level of 0.05, can we conclude that anemia and balanced diet are independent? Apply the Chi Square test to determine if there is a significant association between these two variables in the sample of school children from the rural community. Interpret the results obtained and draw conclusions about the relationship with food in the community studied. Reflection Question: After performing the Chi-Square analysis and obtaining the results, reflect on what other variables could influence reducing the prevalence of anemia that were not included in this study?

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Answer to a math question 1. Case Study: Prevalence of Mild Anemia in School Children in a Rural Community in Peru Introduction: Mild anemia is a public health problem in Peru, especially in rural areas where access to adequate nutrition may be limited. This study seeks to investigate the prevalence of mild anemia in schoolchildren in a rural community in Peru and its possible association with access to a balanced diet. Problem: What is the prevalence of mild anemia in schoolchildren in a rural community in Peru and how is it related to access to a balanced diet? / Determine whether there is a significant association between dichotomous variables. Anemia (Present/absent). Balanced diet (Yes / No). Data were collected from 95 schoolchildren in a rural community in the country, recording their anemia status (present or absent) and their access to a balanced diet (yes or no), as assessed by parents or guardians. The data are presented in the following table: Table 1: Type of diet and anemia in children in a rural community, Peru 2022. Balanced diet Anemia Present Absent No 32 12 Yes 9 42 With a significance level of 0.05, can we conclude that anemia and balanced diet are independent? Apply the Chi Square test to determine if there is a significant association between these two variables in the sample of school children from the rural community. Interpret the results obtained and draw conclusions about the relationship with food in the community studied. Reflection Question: After performing the Chi-Square analysis and obtaining the results, reflect on what other variables could influence reducing the prevalence of anemia that were not included in this study?

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Gene
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108 Answers
1. **Datos Observados:**

- Sin alimentación balanceada:
- Anemia presente: O_{11} = 32
- Anemia ausente: O_{12} = 12

- Con alimentación balanceada:
- Anemia presente: O_{21} = 9
- Anemia ausente: O_{22} = 42

2. **Totales:**

- Filas:
- Sin alimentación balanceada: 32 + 12 = 44
- Con alimentación balanceada: 9 + 42 = 51

- Columnas:
- Anemia presente: 32 + 9 = 41
- Anemia ausente: 12 + 42 = 54

- Total general: 44 + 51 = 95

3. **Valores Esperados (E_{ij}):**

- E_{11} = \frac{44 \times 41}{95} \approx 19
- E_{12} = \frac{44 \times 54}{95} \approx 25
- E_{21} = \frac{51 \times 41}{95} \approx 22
- E_{22} = \frac{51 \times 54}{95} \approx 29

4. **Cálculo de \chi^2:**

- \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
- \chi^2 = \frac{(32 - 19)^2}{19} + \frac{(12 - 25)^2}{25} + \frac{(9 - 22)^2}{22} + \frac{(42 - 29)^2}{29}
- \chi^2 \approx 27.01

5. **Valor P y Grados de Libertad:**
- Grados de libertad = (n_{\text{filas}} - 1) \times (n_{\text{columnas}} - 1) = 1
- Valor P = 0.0000002024

6. **Conclusión:**
- Dado que el valor P es mucho menor que el nivel de significancia de 0.05, se rechaza la hipótesis nula.
- La conclusión es que existe una asociación significativa entre la presencia de anemia y el acceso a una alimentación balanceada.

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Sodium 38.15 38.78 38.5 38.65 38.79 38.89 38.57 38.59 38.59 38.8 38.63 38.43 38.56 38.46 38.79 38.42 38.74 39.12 38.5 38.42 38.57 38.37 38.71 38.71 38.4 38.56 38.39 38.34 39.04 38.8 A supplier of bottled mineral water claims that his supply of water has an average sodium content of 36.6 mg/L. The boxplot below is of the sodium contents levels taken from a random sample of 30 bottles. With this data investigate the claim using SPSS to apply the appropriate test. Download the data and transfer it into SPSS. Check that your data transfer has been successful by obtaining the Std. Error of the mean for your data which should appear in SPSS output as 0.03900.. If you do not have this exact value, then you may have not transferred your data from the Excel file to SPSS correctly. Do not continue with the test until your value agrees as otherwise you may not have correct answers. Unless otherwise directed you should report all numeric values to the accuracy displayed in the SPSS output that is supplied when your data has been transferred correctly. In the following questions, all statistical tests should be carried out at the 0.05 significance level. Sample mean and median Complete the following concerning the mean and median of the data. mean =  mg/L 95% CI:  to  mg/L Based upon the 95% confidence interval, is it plausible that the average sodium content is 36.9 mg/L?      median:  mg/L The median value is      36.9 mg/L. Skewness Complete the following concerning the skewness of the data. Skewness statistic =        Std. Error =  The absolute value of the skewness statistic     less than 2 x Std. Error Therefore the data can be considered to come from a population that is      . Normality test Complete the following summary concerning the formal testing of the normality of the data. H0: The data come from a population that     normal H1: The data come from a population that     normal Application of the Shapiro-Wilk test indicated that the normality assumption     reasonable for sodium content (S-W(  )=  , p=   ). Main test Using the guidelines you have been taught that consider sample size, skewness and normality, choose and report the appropriate main test from the following ( Appropriate ONE ) You have selected that you wish to report the one-sample t-test. H0: The mean sodium content     equal to 36.9 mg/L H1: The mean sodium content     equal to 36.9 mg/L Application of the one-sample t-test indicated that the mean is      36.9 mg/L (t(  ) =  , p =   ). You have selected that you wish to report the Wilcoxon signed rank test. H0: The median sodium content     equal to 36.9 mg/L H1: The median sodium content     equal to 36.9 mg/L Application of the Wilcoxon signed rank test indicated that the median is      36.9 mg/L (z =  , N =  , p =   ).