1) Assuming the following numbers are dollar amounts per treatment with a specific drug, develop a point estimate of the meancost per treatment of the drug. Next, develop a point estimate of the standard deviationof the cost per treatment of the drug.

4376 5578 2717 4920 4495

4798 6446 4119 4237 3814

2) The population mean SAT score is u = 1020 and a population standard deviation is o = 100. What is the probablity that a random sampleof 75 students will provide a sample mean SAT score within 10 of the population mean? Secondly, what is the probabilitya random sample of 75 students will provide a sample mean SAT score within 20 of the population mean?

3) A simple random sample size of 100 is selected from a population with p = .40.

What is the expected value of p”with a line over it”.

What is the standard errorof p”with a line over it”?

Show the samplingdistribution of p”with a line over it”.

What does the sampling distributionof p”with a line over it” show?

4) A simple random sample of size 50 is selected from a population with o = 10. find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate).

the population size is infinite.

the population size is n = 50,000.

the population size is n = 5000

the population size is n = 500

5) The following dataindicate the number of persons who are bullish, neutral, and bearish on the short-term prospects for the stock market:

Bullish 409 Neutral 299 Bearish 291

Develop a point estimate of the following population parameters:

The proportion who are bullish on the stock market

the proportion who are neutral on the stock market

the proportion who are bearish on the stock market.

1) Assuming the following numbers are dollar amounts per treatment with a specific drug, develop a point estimate of the mean cost per treatment of the drug. Next, develop a point estimate of the standard deviation of the cost per treatment of the drug.

4376 5578 2717 4920 4495 4376 5578 2717 4920 4495

4798 6446 4119 4237 3814 4798 6446 4119 4237 3814

The calculations are shown below

The point estimate of mean and standard deviation are the mean and standard deviation of sample

Mean and standard deviation of the sample

X= X 2 =

4,376 19,149,376

5,578 31,114,084

2,717 7,382,089

4,920 24,206,400

4,495 20,205,025

4,798 23,020,804

6,446 41,550,916

4,119 16,966,161

4,237 17,952,169

3,814 14,546,596

Total= 45,500 216,093,620

n=no of observations= 10

Mean= 4,550 =45500/10

variance={summation of X 2 – n(Mean) 2 }/(n-1)= 1007624.44 =(216093620-10*4550^2)/(10-1)

standard deviation =square root of Variance= 1,003.8 =square root of 1007624.44

a point estimate of the mean cost per treatment of the drug= 4,550.0

a point estimate of the standard deviation of the cost per treatment of the drug= 1,003.8

2) The population mean SAT score is u = 1020 and a population standard deviation is o = 100. What is the probablity that a random sample of 75 students will provide a sample mean SAT score within 10 of the population mean? Secondly, what is the probability a random sample of 75 students will provide a sample mean SAT score within 20 of the population mean?

Note: We are using the Normal Distribution Tables for this problem; z denotes z-score

1) Within plus/minus 10 of the population mean …

5 questions on point estimate of mean, proportion and standard deviation, sampling distribution, standard error etc. have been answered.