Measuring a study sample - Coursework Example

Measuring a study sample Reference country Study sample % of MRPs Briant et al (2004) New Zealand657915.45%Ruiter, R (2012)Netherlands21271331.3%Garrido et al (2010)Spain3508351.69 %McDonnell and Jacobs (2002)USA201660.76%Koh et al (2003)Singapore34710.8%Yee et al (2005)Asilomar, CA216912.6%Kalisch et al (2012)Australia163000813.3%Table (1) List of studies used retrospective methods.Process of Measuring a Sample SizeSample size is the number of units of a given population to be studied. Sample size should be large enough so as to increase the accuracy of the data.

When calculating sample sizes, there are some factors which must be taken into consideration. Some of these are: confidence interval, administrative concerns, cost considerations, minimum acceptable levels, sampling method and the viability within the population,.Determining a sample size is very important because when samples are too large they may waste a lot of time, resources and money; on the other hand, when the samples are too small, the results might be accurate.It is, therefore, prudent that we determine a sample size that will give the required values with the minimum error as possible so that the objective of the study can be accomplished.

The procedure below will give us the procedure of calculating the data assuming that the total population is not given. The study will major on the past studies which gave the above presented values.Calculating the Sample SizeWhen calculating the sample size we are going to assume the confidence interval of 95% with a standard error of 5%.We are, therefore, solving for the sample size n95% degree of confidence corresponds to α= 0.05.We must also calculate the standard deviation.:ORσ = √[ ∑(X-Mean)2/N]N=7Mean=6579+2127133+350835+20166+347+2169+1630008=41372374137237 / 7=591033.

85714Respective deviations=6579-591033.857=-584454.857, 2127133-591033.857=1536099.14, 350835-591033.857=-240198.857, 20166-591033.857=-570867.857, 347-591033.857 =-590686.857, 2169-591033.857=-588864.857, 1630008-591033.857=1038974.14.Square and sum the deviations(-584454.857)2+( 1536099.14)2+(240198.857)2+-(570867.857)2+-(590686.857)2+(-588864.857)2+(1038974.14.)234158747990+2359600568000+57695490900+325890110200+ 348910963000+ 346761819800+ 1079467264000 = 4552484964000Divide the sample by less than 1 the value of N because we are dealing with a sample not a population.

4552484964000 / 7-1= 809985618453.476 - Variance.Rounded off Value = 758747494000Square root the total√809985618453.476Standard deviation = 899992.01022Now we have the Standard deviation,With a correspondence of 5% each shaded region in the diagram has an area of α/2= 0.025.The region to the left z ­­α/2 and the right of z = 0 is therefore 0.5-0.025 = 0.475From the table of standard normal z distribution that area corresponds to z which is 1.96.Therefore, n is equal to n= [ z ­­α/2. σ/E]2[1.

96 X 899992.01022/5]235279.6122n=