a. Let p
1
be the proportion of U.S. adults who planned to watch at least a "fair amount" of the 2018 sporting event and p
2
be the proportion of U.S. adults who planned to do so in 2014. We want to test the following hypotheses:
H
0
:p
1
=p
2
H
a
:p
1
<p
2
We will use a significance level of α=0.05.
Our sample size is n=2232 and our sample proportion is
p
^
=0.4086. The standard error of the sample proportion is:
SE(
p
^
)=
n
p(1−p)
=
2232
0.4086(1−0.4086)
=0.0097
Our test statistic is:
z=
SE(
p
^
)
p
^
−p
0
=
0.0097
0.4086−0.47
=−6.38
Our p-value is P(z<−6.38)=0.000000000002.
Since our p-value is less than our significance level of α=0.05, we reject the null hypothesis. There is evidence to suggest that the proportion of U.S. adults who planned to watch the 2018 sporting event was less than the proportion who planned to do so in 2014.
b. A 90% confidence interval for the population proportion is:
CI=(
p
^
−z
1−α/2
⋅SE(
p
^
),
p
^
+z
1−α/2
⋅SE(
p
^
))
CI=(0.4086−1.645⋅0.0097,0.4086+1.645⋅0.0097)
CI=(0.4002,0.4170)
We are 90% confident that the true proportion of U.S. adults who planned to watch at least a "fair amount" of the 2018 sporting event is between 0.4002 and 0.4170.
Our confidence interval does not include 0.47, the proportion of U.S. adults who planned to watch at least a "fair amount" of the sporting event in 2014. This supports our hypothesis test conclusion that the proportion of U.S. adults who planned to watch the 2018 sporting event was less than the proportion who planned to do so in 2014.