Please assist me with this discussion post Based on your course project data, pl
ID: 3170957 • Letter: P
Question
Please assist me with this discussion post
Based on your course project data, please answer the following questions:
What margin of error did you calculate for your estimate of the population parameter?
Suppose you would like to decrease the margin of error to half its actual value (which you calculated in the previous question), what sample size would you need?
Discuss the various issues that you would consider in determining an appropriate sample size for your study.
Study is based on childhood obesity ages 8-17.
Weight BMI Age 21.6577181 8.74496644 9.134228188 Mean 8 8 Median 22 6 6 Mode 22 6 6 Minimum 17 12 27 Maximum 69 6 63 10 Range 9.4158 Variance 4.4075 28.7521 Standard 2.0994 5.3621 3.0685 Deviation 58.70% 14.17% Coeff. of Variation 24.01% 0.2321 Skewness 9.5104 0.1564 Kurtosis -1.3077 -1.0141 106.3054 149 149 149 Count 0.1720 0.4393 0.2514 Standard ErrorExplanation / Answer
What margin of error did you calculate for your estimate of the population parameter?
Here the question is based on sample size.
The formula for sample size is,
n = [(Zc*sigma)/E]2
where n is the sample size
Zc is the critical value for normal distribution.
Zc we can find by using EXCEL.
syntax :
=NORMSINV(probability)
where probability = 1 - a/2
a = 1 - C
C is confidence level.
Sigma is the standard deviation
sqrt(n) = [(Zc*sigma)/E]
E = (Zc*sigma)/sqrt(n)
This is the modified formula for calculating margin of error.
Assume at 95% confidence i.e C = 95%
Zc = 1.96
Now for age variable :
sigma = 2.0994
n = 149
E = (Zc*sigma)/sqrt(n)
= (1.96*2.0994) / sqrt(149) = 0.3371
Similarly for weight variable :
sigma = 5.3621
n = 149
E = (Zc*sigma)/sqrt(n)
E = (1.96*5.3621) / sqrt(149) = 0.8610
And for variable BMI :
sigma = 3.0685
n = 149
E = (Zc*sigma)/sqrt(n)
E = (1.96*3.0685) / sqrt(149) = 0.4927
Suppose you would like to decrease the margin of error to half its actual value (which you calculated in the previous question), what sample size would you need?
Here for each variable we have to half the margin of error and find sample size.
n = [(Zc*sigma)/E]2
For age :
Here E = 0.3371/2 = 0.1686
n = [(1.96*2.0994)/0.1686]2 = 595.998 which is approximately equal to 596.
For weight :
E = 0.8610 / 2 = 0.4305
n = [(1.96*5.3621) / 0.4305]2 = 595.985 which is approximately equal to 596
For BMI :
E = 0.4927/2 = 0.2464
n = [(1.96*3.0685) / 0.2464]2 = 596.0178 which is approximately equal to 596
It means that we get same sample size for three variables by half the margin of error.