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SPA 3e_ Teachers Edition _ Ch 6

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426<br />

C H A P T E R 6 • Sampling Distributions<br />

• The sampling distribution of p^ is less variable for larger samples. This is indicated<br />

by the !n in the denominator of the standard deviation formula.<br />

• The formula for the standard deviation of the distribution of p^ requires that the<br />

observations are independent. In practice, we are safe assuming independence<br />

when sampling without replacement as long as the sample size is less than 10%<br />

of the population size.<br />

Alternate Example<br />

Expensive ride?<br />

Mean and SD of the sampling<br />

distribution of p^<br />

PROBLEM: Suppose that 26% of all<br />

high school students at a large school<br />

spend about half or more of their<br />

earnings on a car. A random sample<br />

of 150 students from this school is<br />

surveyed. Let p^ 5 the proportion of<br />

students in the sample who spend about<br />

half or more of their earnings on a car.<br />

(a) Calculate the mean and standard<br />

deviation of the sampling distribution of p^ .<br />

(b) Interpret the standard deviation<br />

from part (a).<br />

SOLUTION:<br />

(a) m p^ = 0.26 and<br />

0.26(1− 0.26)<br />

s p^ =<br />

= 0.036<br />

Å 150<br />

(b) In SRSs of size n 5 150, the sample<br />

proportion of students who spend about<br />

half or more of their earnings on a car<br />

will typically vary by about 0.036 from<br />

the true proportion of p 5 0.26.<br />

a<br />

e XAMPLe<br />

What proportion of students have a smartphone?<br />

Mean and SD of the sampling distribution of p^<br />

PROBLEM: Suppose that 43% of students at a large high school own a smartphone. As part of a<br />

schoolwide technology study, the principal surveys an SRS of n 5 100 students. Let p^ 5 the proportion<br />

of students in the sample who own a smartphone.<br />

(a) Calculate the mean and the standard deviation of the sampling distribution of p^ .<br />

(b) Interpret the standard deviation from part (a).<br />

SOLUTION:<br />

0.43(1 − 0.43)<br />

(a) m p^ = 0.43 and s p^ = = 0.050<br />

Å 100<br />

(b) In SRSs of size n 5 100, the sample proportion of students<br />

who own a smartphone will typically vary by about 0.050 from<br />

the true proportion of p 5 0.43.<br />

d<br />

ThinK ABoUT iT Is the sampling distribution of p^ (the sample proportion of<br />

successes) related to the sampling distribution of X (the sample count of successes)?<br />

Yes!<br />

number of successes in sample<br />

p^ = = X sample size<br />

n<br />

For example, here are dotplots showing the simulated sampling distribution of X 5 the<br />

number of pennies from the 2000s and p^ 5 the proportion of pennies from the 2000s<br />

for samples of size 20 in Mr. Ramirez’s class. The distributions are exactly the same,<br />

other than the scale on the axis.<br />

d<br />

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6 8 10 12 14 16 18 20<br />

Number of pennies from 2000s<br />

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In this context, n 5 100 and p 5 0.43. The mean is m p^ 5 p<br />

p(1 − p)<br />

and the standard deviation is s p^ 5 . Å n<br />

FOR PRACTICE TRY EXERCISE 1.<br />

d<br />

d d<br />

d d<br />

d d d<br />

d d d d<br />

d d d d d d<br />

d d d d d d<br />

d d d d d d d<br />

d d d d d d d d<br />

d d d d d d d d d d d<br />

0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0<br />

Proportion of pennies from 2000s<br />

Starnes_<strong>3e</strong>_CH06_398-449_Final.indd 426<br />

Common Error<br />

Students may refer to the sampling<br />

distribution of the sample proportion of<br />

pennies as a binomial distribution. Tell them<br />

this is not technically correct because a<br />

binomial random variable is a count, not a<br />

proportion. However, these two distributions<br />

are mathematical transformations of one<br />

another.<br />

18/08/16 5:01 PMStarnes_<strong>3e</strong>_CH0<br />

426<br />

C H A P T E R 6 • Sampling Distributions<br />

Starnes_<strong>3e</strong>_ATE_CH06_398-449_v3.indd 426<br />

11/01/17 3:56 PM

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