Transcript Ch. 3 – Displaying and Describing Categorical Data

Warm Up
The average amount of meat that an American
consumes per year is 218.4 lbs. Assume that the
standard deviation is 25 and that the
distribution of meat eaten by Americans is
approximately normal. If an individual is
selected at random, find the probability that he
or she consumes less than 210 lbs of meat per
year.
Part V – From the Data at Hand to the World at Large
Ch. 18 – Sampling Distribution Models
(Day 2 –Sample Means)
Sample Means
• Yesterday, we examined the sampling
distribution of sample proportions
• Today, we will look at a similar idea – the
distribution of sample means
• We are still imagining the results of repeated
samples, but instead of recording how often a
categorical outcome occurs (proportion), we
are recording the average value (mean) of a
quantitative variable
Individual Values v. Sample Means
• In the warm-up, we calculated the probability that a
randomly selected individual American eats less than
210 lbs of meat per year
• Let’s look at the picture we drew again:
.3669
210
218.4
• What if we had selected 40 Americans at random,
and found the average amount of meat eaten per
year by this group? Do you think this value has a
higher or lower chance of being below 210 lbs?
Individual Values v. Sample Means
• The distribution of sample means has less
variability than the distribution of individual
values
Bigger Sample Sizes
• What do you think would happen if we looked
at a sample of 100 people? 1000?
• As the sample size gets bigger, the distribution
becomes less variable
• However, the average value stays the same
• Mean and standard deviation for a sampling
distribution of sample means:
x  
x 

n
Back to our problem…
• If a random sample of 40 Americans is selected,
what is the probability that the mean amount
of meat eaten by those sampled is less than
210 lbs per year?
statistic  parameter
z
standard deviation
z
210 218.4
210  218.4
z
 2.13
25
40
x  

n
P( x  210)  .0166
One more example…
• The scores of students on the ACT college entrance
examination in a recent year had the normal
distribution with a mean score of 18.6 and a
standard deviation of 5.9.
a) What is the probability that a single randomly
chosen student has a score 21 or higher?
statistic  parameter
z
standard deviation
18.6 21
21  18.6
z
 0.41
5.9
z
x

P( x  21)  1  .6591  .3409
One more example (continued)…
• The scores of students on the ACT college entrance
examination in a recent year had the normal
distribution with a mean score of 18.6 and a
standard deviation of 5.9.
b) What is the probability that the mean score for a
random sample of 50 students is 21 or higher?
statistic  parameter
z
standard deviation
18.6 21
z
x 

n
21  18.6
z
 2.88
5.9
50
P( x  21)  1  .9980  .0020
Conditions for Sample Means
• For today, some of your problems will involve sample
means for normally distributed variables
• We are going to adjust these a little tomorrow, but for
now, you should check:
1) Randomization: The sample must be selected randomly
2) 10% Condition: The sample size must be less than 10%
of the population size
3) Normal Sampling Distribution: The sampling distribution
must be approximately normal. This happens
automatically when the variable being studied is stated
to be normally distributed. We will revisit this 3rd
condition tomorrow.
Assignment 18-2
Pg. 432 # 37, 38