Section 9.2, cont. – Two-sided Tests, Confidence Intervals, Type II Error and Power
Example, p. 562 Nonsmokers According to the Centers for Disease Control and Prevention (CDC) web site,
50% of high school students have never smoked a cigarette. Taeyeon wonders whether this national result
holds true in his large, urban high school. For his AP Statistics class project, Taeyeon surveys an SRS of 150
students from his school. He gets responses from all 150 students, and 90 say that they have never smoked a
cigarette. What should Taeyeon conclude? Give appropriate evidence to support your answer.
State: We want to perform a significance test using the hypotheses
H0: p = 0.50
Ha: p ≠ 0.50
Where p is the proportion of all students in Taeyeon’s school who would say they have never smoked
cigarettes. We’ll use α = 0.05.
Plan: One-sample z test for p
Conditions: 1. SRS of 150 students from his school.
1
10%: 150 ≤ 10 (all students in his high school)
Reasonable to assume
2. Large Counts: 150(0.5) = 75 ≥ 10 and 150(1 – 0.5) = 75 ≥ 10
Safe to use Normal calculations
90
Do: 𝑝̂ = 150 = 0.60, 𝑝0 = 0.50
𝑧=
𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐−𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐
=
𝑝̂−𝑝0
̂
̂
√𝑝(1−𝑝)
𝑛
=
0.60−0.50
0.5(0.5)
√
150
= 2.45
*When looking for the P-value in the chart, it
will only show one side. You have to double
the probability for both sides.
The calculator will automatically do this for you.
P-value = 0.0142
Conclude: Because the P-value of 0.0142 is less than α = 0.05, we reject H0. We have convincing
evidence that the proportion of all students at Taeyeon’s school who would say they have never smoked
differs from the national result of 0.50.
Why Confidence Intervals Give More Information
Example, p. 563 Nonsmokers Taeyeon found that 90 of an SRS of 150 students said they had never smoked a
cigarette. We checked the conditions for performing the significance test earlier. Before we construct a
confidence interval for the population proportion p, we should check that both n𝑝̂ ≥ 10 and n(1 - 𝑝̂ ) ≥ 10.
(150)(0.60) = 90 ≥ 10
and
(150)(0.40) = 60 ≥ 10
We’ll use a 95% confidence interval since we used a significance level of 0.05 (1 – 0.05 = 0.95).
Our 95% confidence interval is:
𝑝̂ (1 − 𝑝̂ )
0.6(0.4)
𝑝̂ ± 𝑧 ∗ √
= 0.60 ± 1.96√
= 0.60 ± 0.078 = (0.522,0.678)
𝑛
150
We are 95% confident that the interval from 0.522 to 0.678 captures the true proportion of students at
Taeyeon’s high school who would say that they have never smoked a cigarette.
The confidence interval gives more information than the two-sided z test because it provides plausible values
for p based off of the sample.
Type II Error and the Power of a Test
Power – The power of a test against a specific alternative is the probability that the test will reject H0 at a
chosen significance level α when the specified alternative value of the parameter is true. (probability of
AVOIDING a Type II error)
Power = 1 – P(Type II error) = 1 - β
We would prefer Power to be high. So what can
we do to reduce the P(Type II error)?



Increase n. More data provides better
information which in turn helps us to
avoid Type II errors.
Increase α. It is easier to reject H0 with a
larger significance level, which means
we have a smaller chance of making a
Type II error.
Increasing the distance of the alternative
from the null.
β
Homework: p. 572 #45, 47, 49, 55, 57 b, 61
Due: Tuesday