Question

Problem 1- Glaucoma is a disease of the eye that is manifested by high intraocular pressure. The distribution of intraocular pressure in the general population is approximately normal with mean 16 mm Hg and standard deviation 3 mm Hg. In a random sample of 60 people, find the probability that the average intraocular pressure is between 15.2 and 16.6. Problem 2 In a study of perception, 118 men are tested and 16 are found to have red/green color blindness. (a) Find a 90% confidence interval for the true proportion of men from the sampled population that have this type of color blindness. (b) Using the results from the above mentioned survey, how many men should be sampled to estimate the true proportion of men with this type of color blindness to within 2% with 99% confidence? (c) If no previous estimate of the sample proportion is available, how large of a sample should be used in (b)?

Answer 1

Problem 1-

X : Intraocular pressure

X follows normal distribution with mean: 16mm Hg and standard deviation 3 mm Hg

By Central limit theorem sampling distribution of sample mean (with sample size n) follows normal distribution with mean and standard deviation : /

: average intraocular pressure (Sample size : n=60) follows normal distribution with mean 16mm Hg and standard deviation

3/ = 0.3873

probability that the average intraocular pressure is between 15.2 and 16.6 = P(15.2 < < 16.6)

P(15.2 < < 16.6)= P( < 16.6) - P( < 15.2)

Z-score for 16.6 = (16.6-16)/0.3873 = 1.55 ; Z-score for 15.2 = (15.2-16)/0.3873 = -2.07

From standard tables, P(Z<1.55) =0.9394 ; P(Z<-2.07) = 0.0192

P(15.2 < < 16.6)= P( < 16.6) - P( < 15.2) = P(Z<1.55) - P(Z<-2.07) =0.9394 - 0.0192=0.9202

probability that the average intraocular pressure is between 15.2 and 16.6 = 0.9202

Problem 2

(a)

90% confidence interval for the true proportion of men from the sampled population that have this type of color blindness.

Formula for confidence interval for single population proportions : p

for 90% confidence level = (100-90)/100=0.10

/2 = 0.10/2=0.05

Z_/2 = Z_0.05 = 1.6449

Number of men test : n= 118

Number of men found to have red/green color blindness : x = 16

Sample proportion men to have red/green color blindness : = 16/118 = 0.1356

90% confidence interval for the true proportion of men from the sampled population that have this type of color blindness.

90% confidence interval for the true proportion of men from the sampled population that have this type of color blindness=(0.0838 ,0.1874 )

(b)

Using the results from the above mentioned survey, how many men should be sampled to estimate the true proportion of men with this type of color blindness to within 2% with 99% confidence

Formula for sample size to estimate the confidence interval for true proportion; using a previous value of proportion :p_g

for 99% confidence level = (100-99)/100=0.01

/2 = 0.01/2=0.005

Z_/2 = Z_0.005 = 2.5758

Margin of error : E = 2/100 =0.02

Number of men should be sampled : sample size : n= 1944

(c)

If no previous estimate of the sample proportion is available, how large of a sample should be used in (b)?

Formula for sample size to estimate the confidence interval for true proportion;when no previous estimate of the sample proportion is available

Sample should be 4147

Number of men should be sampled : sample size : n= 4147