Question

Q1. A survey of 500 adults aged 18 – 29 years of age revealed that 285 chose to eat fast food for dinner at least once in the past week. Find the sample proportion of individuals surveyed who ate fast food for dinner at least once in the past week.

(Hint: p ^ = x n , where x = # of people who have given characteristic (eat fast food) and n is sample size)

Q2. Verify two conditions ( a ) n < 0.05 N ( b ) n p ( 1 − p ) ≥ 10 to find whether the sampling distribution is approximately normal. And determine μ p ^ and σ p ^ from the given parameters. Assume the size of the population is 25,000.

n = 200  p = 0.75

Q3. Sixteen percent of Americans do not have health insurance. Suppose a simple random sample of 500 Americans is obtained. In a random sample of 500 Americans, what is the probability that more than 20% do not have health insurance?

[Hint: Find P ( p ^ > 0.20 )]

Q4. Home sizes in Anytown, USA have a mean of 2400 square feet and a standard deviation of 450 square feet. What is the probability that a random sample of 50 homes in Anytown, USA has mean square footage less than 2200 square feet?

Hint: Find P (x ¯ < 2200), and make sure you use μ x ¯ and  σ x ¯ as a mean and SD in normalcdf function)

Q5. A simple random sample of size n = 40 is obtained from a population with μ = 50 a n d σ = 4. Does the population distribution need to be normally distributed for the sampling distribution of to  x ¯ be approximately normally distributed? Why or why not? What is the mean and standard deviation of the sampling distribution?

What if the random sample of size n = 10? Does the population distribution need to be normal? Why or why not?

Answer 1

Q1. A survey of 500 adults aged 18 – 29 years of age revealed that 285 chose to eat fast food for dinner at least once in the past week. Find the sample proportion of individuals surveyed who ate fast food for dinner at least once in the past week.

(Hint: p ^ = x n , where x = # of people who have given characteristic (eat fast food) and n is sample size)

So,

Answer is:

0.57

Q2. Verify two conditions ( a ) n < 0.05 N ( b ) n p ( 1 − p ) ≥ 10 to find whether the sampling distribution is approximately normal. And determine μ p ^ and σ p ^ from the given parameters. Assume the size of the population is 25,000.

n = 200  p = 0.75

(a)

n = 200 < 0.05 X 25,000 = 1250. So, this condition is satisfied.

(b)

n p ( 1 - p) = 200 X 0.75 X 91 - 0.75) = 37.5 10. So, this condition is satisfied.

So,

the sampling distribution is approximately normal.

Q3. Sixteen percent of Americans do not have health insurance. Suppose a simple random sample of 500 Americans is obtained. In a random sample of 500 Americans, what is the probability that more than 20% do not have health insurance?

[Hint: Find P ( p ^ > 0.20 )]

p = 0.16

n = 500

To find P( > 0.20):

Z = (0.20 - 0.16)/0.0164

= 2.4398

By Techology, Cumulative Area Under Standard Normal Curve = 0.9927

So,

P( > 0.20):= 1 - 0.9927

                = 0.0073

So,

Answer is:

0.0073

Q4. Home sizes in Anytown, USA have a mean of 2400 square feet and a standard deviation of 450 square feet. What is the probability that a random sample of 50 homes in Anytown, USA has mean square footage less than 2200 square feet?

Hint: Find P (x ¯ < 2200), and make sure you use μ x ¯ and  σ x ¯ as a mean and SD in normalcdf function)

= 2400

= 450

n = 50

SE = 450/

= 63.6396

To find P( < 2200):

Z = (2200 - 2400)/63.6396

= - 3.1427

By Techology, Cumulative Area Under Standard Normal Curve = 0.0008

So,

P( < 2200): = 0.0008

So,

Answer is:

0.0008

Q5. A simple random sample of size n = 40 is obtained from a population with μ = 50 a n d σ = 4. Does the population distribution need to be normally distributed for the sampling distribution of to  x ¯ be approximately normally distributed? Why or why not? What is the mean and standard deviation of the sampling distribution?

What if the random sample of size n = 10? Does the population distribution need to be normal? Why or why not?

Since the population standard deviation is provided, the population distribution need not be normally distributed for the sampling distribution of to  x ¯ be approximately normally distributed because by Central Limit Theorem, since Sample Size = n = 40 > 30 Large Sample, the sampling distribution of sample mean is normal distribution irrespective of the shape of the population distribution.

the mean of the sampling distribution = 50

standard deviation of the sampling distribution = 4/ = 0.6325

iI the random sample of size n = 10, the population distribution need not be normal because the population standard deviaion is provided and so Z statistic can be used instead of t distribution.