Download of the sample that say they drink the cereal milk.

Sample Proportions
Choose an SRS of size n from a large population with population proportion p having some
characteristic of interest. Let 𝑝̂ be the proportion of the sample having that characteristic.
Then:
The MEAN of the sampling distribution is exactly p.
The STANDARD DEVIATION of the sampling distribution is √
𝑝(1−𝑝)
𝑛
NOTE: Use the formula for the standard deviation of 𝑝̂ only when the population is at least 10
times as large as the sample.
We will use the normal approximation to the sampling distribution (i.e. normalcdf) of 𝑝̂ for
values of n and p that satisfy np ≥ 10 and n(1-p) ≥ 10.
Try this example:
A poll asked a random sample of 912 US adults what they do with the milk in the
bowl after they have eaten the cereal. Of the respondents, 68% said that they drink
it. Suppose that 70% of US adults actually drink the cereal milk.
a. Find the mean and standard deviation of the proportion 𝑝̂ of the sample that say they
drink the cereal milk.
b. Explain why you can use the formula for the standard deviation of 𝑝̂ in this setting.
c. Check that you can use the normal approximation for the distribution of 𝑝̂ .
d. Find the probability of obtaining a sample of 912 adults in which 68% or fewer say they
drink the cereal milk. Do you have any doubts about the result of this poll?
e. What sample size would be required to reduce the standard deviation of the sampling
proportion to one-half the value you found in a.
Solution:
a. By the formulas from above, the mean = 0.7 and the SD = √
.7(1−.7)
912
= 0.0152
b. The sample size is 912 and the population of US adults is greater than 10 times that
amount.
c. We must check that np ≥ 10 and n(1-p) ≥ 10:
912(0.7) = 638.4 and 912(0.3) = 273.6
Since both of these values are greater than 10, we can use normal
approximations.
d. P(p≤0.68) = normalcdf(-999,0.68,0.7,0.0152) = 0.094
So about 9.4% of the time we should expect to obtain a sample of 912
adults in which 68% or fewer say they drink the cereal milk which is
unusual, but not unheard of. I do not doubt the results of the poll.
e. The formula is SD = √
𝑝(1−𝑝)
𝑛
. If we want the left side of the equation (SD) to be
divided by 2, we must also divide the right side by 2. To put that in a square root would
be 4. So the n must be multiplied by 4. Our sample size would need to be 912*4 = 3648
to reduce our standard deviation by half.
You could also find that by solving this equation: (note ½(0.0152) = 0.0076)
0.0076 = √
.7(1−.7)
0.00005776 =
𝑛
square both sides
.7(1−.7)
𝑛
0.00005776n = 0.21
n = 3635.7 Or a sample size of 3636 (There is a slight difference b/c the 0.0152 value
was rounded – it’s close enough though.)
Try another example of the next page.
A poll on exercise habits was asked of 1500 adults. Suppose that 16% of all adults jog.
a.
b.
c.
d.
e.
Find the mean and standard deviation of the proportion 𝑝̂ of the sample who jog.
Explain why you can use the formula for the standard deviation of 𝑝̂ in this setting.
Check that you can use the normal approximation for the distribution of 𝑝̂ .
Find the probability that between 14% and 18% of the sample jog.
What sample size would be required to reduce the standard deviation of the sample
proportion to one-third the value you found in a?
SOLUTION:
a. Mean = .16 and SD = √
.16(1−.16)
1500
= 0.0095
b. The adult population is greater than 1500*10.
c. 0.16 * 1500 = 240 and 0.84*1500 = 1260. Both are greater than 10 so a normal
approximation is safe.
d. P(0.14 ≤ p ≤ 0.18) = normalcdf(0.14,0.18,0.16,0.0095) = 0.9647
e. We would need to multiply the sample size by 9. So 1500 * 9 = 13,500.