At a certain college, 53% of employees carry dependent family members on their h

Question

At a certain college, 53% of employees carry dependent family members on their health insurance plans. Currently, there are 175 employees working for the college. A random sample of 60 employees is selected. Complete parts a through c below. (round answers to 4 decimal places)

a. What is the probability that less than 58% of the sample carries dependent family members on their health insurance plan?

b. Answer the question in part a using sample sizes of 100 and 150. Explain the differences in these probabilities. First find the probability for a sample size of 100.

Find the probability for a sample size of 150.

Explain the differences in these probabilities.

c. Answer the question in part a without the finite correction factor. Explain the differences in these probabilities.

Explain the differences in these probabilities.

Explanation / Answer

Let N =size of population = 175

n = size of sample = 60

p = 53% or .53

standard deviation () = (p*q)/n * ((N-n)/(N-1)

Where p = proportion

q= 1- p = 1 -0.53 = 0.47

(N-n)/(N-1) = finite population correction

Hence = [(.53*.47)/60] * (175-60)/175-1) = 0.064433 * 0.81297 = 0.05238

Now as the sample size = 60>30, we apply the central limit theorem and consider the sample to follow normal distribution

Part a) To find probability that less than 58% of sample carries dependent family members

In order to find the probability we first need to find the z score corresponding to phat = .58

Hence Z = (phat - p) /

= (.58 - .53) / 0.05238

= 0.95456

Hence Z = 0.9545

Look in to the Z table to find the area under the curve for the Z value 0.9545

P(Z<0.9545) = 0.8289

Part b) find probabilty for sample size =100

= (.53*.47)/100 * (175-100)/175-1) = 0.0328

Z = (.58-.53) / .0328 = 1.524

P(Z<1.524) = 0.9357

Now find probability for sample size (n) = 150

= (.53*.47)/150 * (175-150)/(175-1) = 0.01544

Z = (.58 - .53) / 0.01544 = 3.24

P(Z<3.24) = 0.9994

Conclusion: for sample size 60 probability is 0.8289;

for sample size 100 probability is 0.9357

for sample size 150 probability is 0.9994

We observe that as the sample size increases, the standard error decreases and the probability of the proportion to be less that 58% increases.

Part c) If we remove the finite correction factor((N-n)/(N-1) for the part a) in the calculation of standard deviation

we get = (p*q)/n

= (.53*.47)/60

= 0.064433

Z = (.58-.53)/.064433

= 0.7760

P(Z<0.776) = 0.7811

Conclusion: with finite correction factor the probability is 0.8289

without finite correction factor the probability is 0.7811

Hence we observe that with finite correction factor for small samples(like n=60) the probability is more as compared to the probability obtained without finite correction factor. The logic behind increase in probability is that for small samples with the use of finite correction factor standard error gets reudced and hence probability increases.

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