a.
Determine the value of the sample proportion
of Californians who have visited Yosemite. Is
𝑝
^
this value higher or lower than the reported population value?
= 240 /1000 = 0.24
𝑝
^
This value is lower than the reported population value.
b.
Determine the expected value of the sample proportion
.
µ
𝑝
^
=
= 0.28
µ
𝑝
^
𝑝
c.
Determine the standard deviation of the sample proportion
.
σ
𝑝
^
=
= 0.0142
σ
𝑝
^
0.28(1−0.28)
1000
d.
Determine that the condition for normality is satisfied.
1000(0.28) = 280 & 1000(1-.28) = 720
Both values are greater than 10, therefore the conditions for normality are satisfied.
e.
Determine the probability the sample proportion exceeds 0.24.
P(
> 0.24) = P(Z >
) = P(Z > -2.82) = 1-0.0024 = 0.9976
𝑝
^
0.24−0.28
0.0142
P(x > 0.24) = 0.9976
f.
Determine the probability the sample proportion is between 0.2 and 0.3.

P(0.2<x<0.3) = 0.9205
Formulas for Confidence Intervals
Confidence interval for
μ
when
σ
is known
Use
Inverse Normal Table
- between
𝑋
‾
± 𝑍 ·
σ
𝑛
Confidence interval for
μ
when
σ
is unknown
Use
Inverse t table
df=n-1
𝑋
‾
± ? ·
?
𝑛
4.
20 students were asked how many units they were taking at a community college.
5
5
5
8
9
10
10
13
15
15
15
15
15
16
18
19
20
21
22
24
a.
Calculate the sample mean. This is a
point estimator
for the population mean,
μ
. Explain
what this means.
=
14. This means that the average population taking courses at the community
𝑋
‾
college is taking 14 units.
b.
Find a 95% confidence interval for the population mean,
assuming σ = 5
. Calculate and
explain the margin of error
Confidence interval = 2.19 Margin of error= 16.19 and -11.81
14(+-) 1.96 X
5
20
2.19 + 14= 16.19