Sample Means:

Target Goal: I can find the mean and standard deviation of the sampling distribution.

7.3a

h.w: pg 441:43 – 46; pg 454: 49, 51, 53, 55

x

Because sample means are just averages of observations, they are among the most common statistic.

• Bull Market or Bear Market

• Rates of return for NYSE in 1987.• What do you notice compared to the

next “portfolio”?

Rates of return for NYSE in 1987

• Data is wide spread and variability is high.

Rates of return for “5 stock portfolios”

• Data is less spread with less variation.

Sample means are:

Averages are more normal than individual observations.

Averages are less variable than individual observations.

The Mean and Standard Deviation of the Sample Mean

• Suppose that is the mean of an SRS of size n drawn from a large population with mean μ and standard deviation σ.

xx

,x x

n

• Note: We can only use when

the population is at least 10 times the sample.

(This is almost always the case.)

n

The Shape of the distribution of depends on the shape of the distribution.

• If drawn from a population that is normal then:

Sample mean has:

x

x

: ,x Nn

• Note: the sample mean will not be as exact as the population mean because of sampling variability.

Example : Young Woman’s Heights

• The heights of young women vary according to N(64.5, 2.5).

• This is the distribution of one individual chosen at random.

• Take a SRS of 10 women.

Find the Mean and Standard Deviation

= 64.5

The average height of individual women vary widely (2.5inches) about the population mean.

• But, the average height of a sample of 10 women is less variable (.79 inches).

x

x

n

0

2.5

1

0.79x inches

What if n = 100?

= 0.25

The average height of a sample of 100 women is has even less variable.

2.

100

5x

What is the probability that a randomly selected woman is taller than 66.5 inches?

• Let x = the height of a randomly selected women.

• This follows a N(64.5,2.5)

We want: P(X > 66.5)

Standardize: P(X > 66.5)

= 0.80

= 1 – normcdf(-E99,0.80)

= 1 - .7881

= 0.2119

xz

66.5 64.5

2.5

What is the probability that a SRS of 10 women has a mean height greater than 66.5?

• We want : P( > 66.5) for n = 10

• We know that

: 64.5,.79x N

x

Standardize Sample Mean:

z = 2.53

P(x > 66.5) = P(Z>2.53)= 1 – P(Z≤2.53)= 1 – normcdf(-

E99,2.53)= 0.0057

Or normcdf(2.53,E99)

66.5 64.52.5

10

The probability is much less likely (less than 1% chance) compared to the individual probability which we calculated to be approx, 21% to have a height > 66.5.

1% 21%

Conclusion:

• The averages of several observations are less variable than individual observations.

• So, the sample mean is less variable than a single measurement.

x

• Receive CLT hand out and discuss tomorrow.

• Work on carnival calculations or hw the rest of class.