Study Design, Statistical Hypotheses, And Analysis Methods

Question 1

1. A study was conducted on self-reported patients of osteoarthritis. The study diagnosed the chosen individuals and noted who used CAM or complementary and alternative medicines in addition to or instead of conventional analgesic treatment and whether they had other ailments or not. Additionally it recorded whether they came from low education or high educational backgrounds. The study here only focused on the data that was reported by the sampled individuals at one point of time. It is therefore a cross sectional study design (Pagano  and Gauvreau  2018).
2. b) Considering only the association between educational background and use of CAM, the hypothesis of interest for the given study can be written as follows:

H₀: Education is not associated with CAM use

Against

H₁: Education is not associated with CAM use

OR

H₀: p_ij0 = p_i00.p_0j0

Against

H₁: not H₀

Here p_ijk is the probability of an individual belonging to the (i, j, k)^th cell of the given contingency table, where 0<i<=2 denotes index of the CAM classes, 0<j<=2 denotes the index of the education classes and 0<k<=2 denotes the index of the comorbidity classes. Then marginal probability over the comorbidities is given by, p_{ij0 =}  and marginal over the comorbidities and the education classes is, p_{i00 =}   and the marginal over comorbidities and CAM is p_{0j0 =}  (Le  and Eberly  2016).

1. c) The marginal contingency table over comorbidities is then given as:

The hypotheses can be tested by use of Chi-square test for association. The Chi-square statistic was computed as:  = 0.81366. Then taking alpha as 0.05, the p-value for the chi-square test given by P(

Therefore the test fails to reject null hypothesis and concludes that there is not enough evidence to accept the alternative that there is association between education and CAM use.

d)

The following tables gives the partial tables for comorbidities= “Yes”

Then the odds ratio for a woman who use CAM and have comorbidity given that she is from a highly educated background is. Then the woman is more likely to be highly educated when using CAM for the group with comorbidity.

The following tables gives the partial tables for comorbidities= “No”

Then the odds ratio for a woman with MI who use CAM and do not have comorbidity given that she is from a low education background is. This implies that a woman using CAM with no comorbidities is less likely to be from a high education background.

e)

Considering the conditional association between educational background and use of CAM along with the presence or absence of comorbidities, the hypothesis of interest for the given study can be written as follows:

H₀: Education is not associated with CAM use when controlling for comorbidities

Against

H₁: Education is not associated with CAM use when controlling for comorbidities

OR

H₀: p_ijk= p_i0k.p_0jkfor all k

Against

H₁: not H₀

Here p_ijk is the probability of an individual belonging to the (i, j, k)^th cell of the given contingency table, where 0<i<=2 denotes index of the CAM classes, 0<j<=2 denotes the index of the education classes and 0<k<=2 denotes the index of the comorbidity classes. Here, p_{i0k =} and p_{0jk =}

Then using the Mantel-Haenszel test for conditional independence, the test statistic was computed as:  where =  which is the expected frequency for  and. Here is the frequency of the (i, j)^th cell of the k^th partial table.Then  =   = 10.91 + 9.21= 20.12.

M= = . Here M follows chi squared statistic with degrees of freedom equal to 1. Then the p-value is P( M> 2.61) =1- P( M<=2.61) = 0.1061 > 0.05. Therefore not enough evidence could be found to reject the hypothesis that no association exists under influence of comorbidities. Thus it is concluded that education and CAM use do not have significant association even when controlling for presence of comorbidities.

1. The null and alternate hypothesis for the ANOVA test can be written as :

Question 2

H₀: m _Democrats = m _Republicans = m _Independents   (here, m _group = mean score of group)

Against

H₁: at least one inequality in H₀

1. b)

The boxplot diagram given implies that the support for democrats is greater than that of the Republicans. This is because the minimum value of the democrats is greater than the median score for the Republicans. However for the Republicans and the Independents, no difference could be identified. This is because the maximum measure of support for Republicans is greater than that of the independents but at the same time, the minimum measure for Republicans is lower than that for the Independents (Silvey  2017). Therefore no significant difference could be discerned.

The mans table shows the mean of the measure of support for each political group, namely, democrat, republican and independent respectively. The average support for democrats was 8.36 with standard deviation 1.689, followed by the independents who had an average of 6.76 with standard deviation 1.091 and the republicans have mean support of 6.16 with standard deviation 2.691.  

The ANOVA table then gives the measure of the within  group variation given by the sum of the sum of squared deviation of each point in a particular group from the group mean for all the groups, equal to 122.519 and the between group variation given by the sum of squared deviation of group means from the grand mean. The degree of freedom for the between group is 2 since there are 3 groups and it is 33 for within group because, it is  there are a total of 35 observations. Then the F statitsic was computed to be 3.96 and the p-value for the right sided test of ANOVA was found to be 0.026 which is less than the level of significance 0.05, implying that the null ought to be rejecetd in favour of the alternative (Silvey 2017). This means that the support for the groups vary significantly among one another.

1. d) The ANOVA test is used to compare among the means of multipe or more than two groups on the same variable metric. This is more efficent than doing multiple pair wise comparisons t-tests, since the single comparison procedure of ANOVA uses a general alternative which is often enough to provide meaningful information. This case deals with support to the three political parties and the test rejects the conjecture of them being equal, answering the question whether the election would be neck to neck or not. The test can however now follow up on post hoc t-tests to see which party has more support if the scope of inquiy allows.

1. It is given that out of 300 citizens chosen randomly from a city, 225 are immunized. Then the estimated proportion of immunized people in the city is 225/300 which is equal to 0.75.
2. The sample size required for a test of proportion with precision 95% of being in the range ± 3% of the true proportion is given as (Silvey 2017). Now estimated   where p =0.75 as estimated in previous section and  Then the minimum sample size required is n= = 240100.

It is given that 10 percent of women of childbearing age having myocardial infraction are users of oral contraception. Then the proportion of women with MI who use OC given by p₁ = 0.1. The relative risk(RR) of MI among OC users was given to be 1.8.

Let p₁ = proportion of women with MI who do not use OC

Then as per definition of relative risk, RR = p₁ / p₀.

Then 1.8 = 0.1/ p₀, implying p₀ = 0.055.

Then define =  = 0.077 and = 1- = 0.922

Then as per Chow et al. (2017), the sample size is given by the formula when power is 80%(  = 0.8) and the confidence is 95%( = 1-0.95) is,

N=  

  =  = = 553.28

Therefore the minimum sample size required is 554. The number of control subjects to be taken is thus 554/2 which is 277. The number of case subjects should be 227.

Assuming that the ratio of case to control is r=1, the sample size for the study for power 90% and confidence level 95%, can be computed using the formula (Chow et al. 2017):

N= , where the standard deviation computed from the previous studies is 84 hours and d= difference in ventilation hours = 36 hours.

Again,  and .

So, N= .

Then, the minimum sample size required for the new trail is determined to be 115.

Reference

Chow, S.C., Shao, J., Wang, H. and Lokhnygina, Y., 2017. Sample size calculations in clinical research. Chapman and Hall/CRC.

Le, C.T. and Eberly, L.E., 2016. Introductory biostatistics. John Wiley & Sons.

Pagano, M. and Gauvreau, K., 2018. Principles of biostatistics. Chapman and Hall/CRC.

Silvey, S.D., 2017. Statistical inference. Routledge.

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