Question

1. According to the American Heart Association, 48% of Americans have some form of heart disease (including high blood pressure). A physician randomly samples 200 people and 49.5% of those sampled have some form of heart disease.

a. Which value above is a parameter and which is a statistic?

b. If tomorrow you were going to sample 600 people, around what value do you think the sample percentage will be? Do you think it will be exactly that value?

2. Market research has shown that 21% of the population list pizza as their favorite food.

a. If you randomly select 500 people and will compute the sample proportion that list pizza as their favorite food, what is the mean of the sample proportion?

b. If you randomly select 500 people and will compute the sample proportion that list pizza as their favorite food, what is the standard deviation of the sample proportion?

c. What interval will contain the sample proportion of people that list pizza as their favorite food for 95% of samples of 500 randomly selected people?

3. The average amount of credit card debt is $5,700 with a standard deviation of σ= $1,200.

a. If you are going to select 42 people and compute the sample mean, what is the mean of the sample mean?

b. If you are going to select 42 people and compute the sample mean, what is the standard deviation of the sample mean?

c. If you wanted the standard deviation of the sample mean to be $100, how many people would you have to sample?

Answer 1

1. a) According to the American Heart Association, 48% american havesome form of heart disease. This 48% relates to the entire population of USA. Since, this 48% denotes a property of the entire population a of USA, this is a parameter. So, the parameter value is 48%

A physician selects a random sample of size 200 and found that 49.5% have some form of heart disease. This 49.5% denotes a property of the sample obtained by that particular physician. So, this 49.5% is a statistic.

b) If, tomorrow, i sample 600 people, it is expected that around 48% of the sample will have some form of heart disease, since 48% is the population average. But, since im drawing a random sample, it can be any value. So, i draw a another sample on some other day, i may get a different number. so, i cant expect same exact value each time i sample.

2)

21% of the population list pizza as their favorite food.

So, here, population proportion p=0.21

a) We have drawn a random sample of 500.

Let p̅ be the proportion of people that list pizza as their favourite food. Here, p̅ is the sample proportion = x/n, where x= no of people out of the 500 people that list pizza as their favourite food and n=sample size = 500.

Since, sample mean is an unbiased estimator of population mean, so, mean of sample proportion = µ_p̅ = p = 0.21

b) The standard deviation of sample proportion is = σ_p̅ = sqrt([p*(1-p)]/n) = sqrt([0.21*(1-0.21)]/500) = 0.0182

c) The 95% confidence interval is given as

Where , z= critical value of standard normal distribution at 5% level of significance, this can be obtained from any standard normal table.

z= 1.96

so, the confidence interval is = (0. 174,0.245)

3)

Given, µ= population mean = 5700, σ=population sd = 1200.

a) Given, sample size = n = 42

Let, x̅= sample mean.

Then, E(x̅)= µ = 5700

Since, sample mean is an unbiased estimator of population mean.

b) Var(x̅)= variance of sample mean= (σ^2)/n = (1200*1200)/42

Sd of sample mean = sqrt[(1200*1200)/42] = 185.164

c) Given σ/n = 100

To find n

σ/n = 100

=>1200/n=100

=> n= 12

so, there will be 12 people in the sample.