i) Is the adjusted estimate likely to be closer to or further away from the null value than the unadjusted estimate from part (b)(ii)? Give a brief explanation for your answer. 


1. 200 children aged under 5 years were sampled from a village in Malawi using simple random sampling. All of the children were tested for malaria using a rapid diagnostic test. 51 of the children tested positive. Estimate the prevalence of malaria in children under 5 years old in the village and provide a 95% confidence interval for the estimate. 


Prevalence/ proportion (p) = 51/200 = 0.255

Confidence interval for prevalence = p + Z_α * S.E. where S.E is the standard error of the proportion

S.E. = √(p(1-p)/n) = √(0.255*0.745)/200 = √0.000949875 = 0.0308

Critical value of Z_α statistic at 95% level of confidence is 1.96.

Lower limit = 0.255 – (1.96*0.0308) = 0.194632

Upper limit = 0.255 + (1.96*0.0308) = 0.315368

I am 95% confident that malaria prevalence is greater or equal to 0.194632 and less or equal to 0.315.

This needs to be rounded to the nearest number, I didn’t want you to add more digit, I just wanted you to break it down more, i.e put more space, I wanted the presentation to look clear, like the one in the power point, the same diagram

1. 40% of children had slept under a bed net the night before the survey. Of these 15% tested positive for malaria.
   • (i) Construct a carefully labeled two-way table for the data derived from the 200 children. 


• (ii) Estimate the odds ratio, risk ratio and risk difference for those who did not sleep under a bed net the night before compared to those that did.

Odds ratio (those who did not sleep under net) = (39/12)/ (81/68) = 2.73

Odds ratio (those who slept under net) = (12/39 = 0.30769231)/ (68/81= 0.83950617) = 0.37

Risk ratio for those who did not sleep under net (R1) = 39/51 = 0.76

Risk ratio for those who slept under net (R2) = 12/51= 0.24

Risk difference = R1-R2 = 0.76-0.24 = 0.52

Risk difference is 39/120-12/80 =0.175 (can you check if this is correct)

• (iii) A statistician performed a χ2 test and obtained a p-value of 0.005. Interpret this result.

• iii) χ2 is a Chi-square test statistic that tests for the significance of the association between variables. Using a significance level of 5% the test statistic is significant since the p-value, which is the probability of rejecting the null hypothesis when it is true is less than the significance level(Helio, et al., 2015). Therefore, the null hypothesis will be rejected. This  then concludes that there is a significant relationship between sleeping under a net and contracting malaria. (what is the relationship?)

1. It has been suggested that families of higher socio-economic status are more likely to own bed nets and are also more likely seek malaria treatment when they are ill. Families have been grouped into three categories of socio economic status.
   • (i) Suggest a statistical technique that would let us examine the association between bed net use and malaria infection taking into account the
   • differences in socio economic status. 


• C 1) In order to determine the association between bed net use and malaria infection, we would use the Mantel-Haenszel test. This is because the data would be stratified and Mantel-Haenszel test would be necessary to control for confounding variable (Alexander, et al. 2015).
  • (ii) Which of the three measures of association mentioned in part (b) (ii) will this technique produce?
• The Mantel-Haenszel statistic produces odds ratio. The odds ratio produces for the test is then adjusted to control for the confounding variables in the data (Mantel-Haenszel test).

(iii) Is the adjusted estimate likely to be closer to or further away from the null value than the unadjusted estimate from part (b)(ii)? Give a brief explanation for your answer.

The adjusted estimate is closer (are you sure its closer and not further away) to the null value relative to the unadjusted estimate. This is because the unadjusted estimate is exaggerated by the effect of chance and the confounding variables which leads to bias in the adjusted value. Adjusting this estimate reduces the effect of these confounding variables and chance and reduces the difference between the estimate and the null value (Portney & Watkins, 2013).