#### Rugrats3

##### New Member I recently began this much dreaded course. I am having a difficult time grasping the concepts. I have an exam on Monday September 26, and desparately need help on 2 problems:

Chapter Title is Sampling Methods and the Central Limit Theorem:
We have to use the Z (area's under the normal curve) and student's t distribution charts. These are very confusing to read and find the data in the body of the chart.

Question #1 is:

A population of unknown shape has a mean of 75. You select a sample of 40. The standard deviation of the sample is 5. Compute the probability the sample mean is:

a) Less than 74
b) Between 74 and 76
c) Between 76 and 77
d) Greater than 77

According to a study, it takes 330 minutes for taxpayers to prepare a 1040 tax form. An agency selects a random sample of 40 taxpayers and finds a standard deviation of the time to prepare a form 1040 is 80 minutes.

a) What assumptions do you need to make anout the shape of the population?
a) What is the standard error of the mean in this example?
b) What is the likelihood the sample mean is greater than 320 minutes?
c) What is the likelihood the sample mean is between 320 and 350 minutes?
d) What is the likelihood the sample mean is greater than 350 minutes?

#### quark

Hi Rugrats3,

Welcome! The Z and t tables are not that hard to read once you get used to them. Here is a nice tutorial on how to read the normal table:

http://www.mis.coventry.ac.uk/~styrrell/pages/ntabexam.htm

I'll do part a of #1, the rest of #1 and #2 are very similar.

Let mu and sd be the population mean and standard deviation.
mu = 75
sd = 5

sample size n=40

Let Xbar be the sample mean, se_Xbar be the standard error of sample mean.
se_Xbar
=sd/sqrt(n)
=5/sqrt(40)
=0.791

P(Xbar < 74)
=P[(Xbar-mu)/se_Xbar < (74-mu)/se_Xbar]
=P[Z < (74-75)/0.791]
=P(Z < -1.26)
=1-P(Z<1.26)
=1-0.8962
=0.1038

#### Rugrats3

##### New Member
Thank you for the tutorial, but I'm still confused on finding data that is 0<Z,0.53 for example. And for the probability of exceeding Z like Z is 45%.

#### quark

P(0<Z<0.53)=P(Z<0.53)-P(Z<0)
=0.7019-0.5
=0.2019

You want to look up the z-values from the first column and row of the normal table, then the probability is in the body of the table.

#### Rugrats3

##### New Member
Problem #2

Thank you for the prompt reply. In problem #2, I am lost on the first two parts to the question.

What do they mean by the standard error?

Was the "xbar" portion of the answer meant for problem #2 part a)?

What do they mean by assumptions about the shape of the population?

#### quark

Standard error of the mean is standard deviation divided by square root of sample size. It's the se_Xbar in #1a.

What I did was for 1a only. The question is about sample mean and we calculated P(Xbar < 74).

We assume that the population is normally distributed. You probably know the shape of normal distribution...

#### Rugrats3

##### New Member
Problems 1 and 2

I am really sorry. I wish I could grasp this information better, but its very difficult for me. On problem # 2, which formulas should I use? Is this a case of my being confused because the words take the place of numbers? Please redirect me! Thank you for your patience!

#### quark

Hi, it's okay. statistics is somewhat hard to get in the beginning. #2 is virtually the same problem as #1, 2c can be solved as follows:

Let mu and sd be the population mean and standard deviation.
mu = 330
sd = 80

sample size n=40

Let Xbar be the sample mean, se_Xbar be the standard error of sample mean.
se_Xbar
=sd/sqrt(n)
=80/sqrt(40)
=12.65

P(Xbar > 320)
=P[(Xbar-mu)/se_Xbar > (320-mu)/se_Xbar]
=P[Z > (320-330)/12.65]
=P(Z > -0.79)
=P(Z<0.79)
=0.7852

Hope this helps. #### Rugrats3

##### New Member
New Question

This one is really confusing:

Information indicates the mean amount of life insurance per household is $110,000. This distribution is positively skewed. The standard deviation of the population is unknown. a. A random sample of 50 households revealed a mean of$112,000 and a standard deviation of $40,000. What is the standard error of the mean? b. Suppose you selected 50 samples of households. What is the expected shape of the distribution of the sample mean? I still don't understand "what is the shape..." or how to find an answer with one part being an unknown. Please help! #### JohnM ##### TS Contributor Rugrats3 said: This one is really confusing: Information indicates the mean amount of life insurance per household is$110,000. This distribution is positively skewed. The standard deviation of the population is unknown.

a. A random sample of 50 households revealed a mean of $112,000 and a standard deviation of$40,000. What is the standard error of the mean?

b. Suppose you selected 50 samples of households. What is the expected shape of the distribution of the sample mean?

a. standard error of the mean is the standard deviation divided by the square root of the sample size

SEm = SD / sqrt(n) = 40000/sqrt(50) = 5656.85

b. if you take 50 SAMPLES and take the mean of each sample, the shape of the distribution of sample means will be normal (if the sample size is "large enough")

#### macro90

##### New Member
What all these questions are pretty much trying to illustrate is the Central Limit Theorem.

The population distribution can have any shape but take many samples and the distribution of the sample will be normal.