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2018 poll of 2232 randomly selected U.S. adults found that 40.86% planned to watch at least a "fair amount" of a particular sporting event in 2018.

In 2014, 47% of U.S, adults reported planning to watch at least a "fair amount."
a. Does this sample give evidence 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? Use a 0.05 significance level.
b. After conducting the hypothesis test, a further question one might ask is what proportion of all U.S. adults planned to watch at least a "fair amount" of the sporting event in 2018. Use the sample data to construct a 90% confidence interval for the population proportion. How does your confidence interval support your hypothesis test conclusion?

User Sam Bobel
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Final answer:

In part a, a hypothesis test can be performed to determine if the proportion of U.S. adults who planned to watch the 2018 sporting event was less than in 2014. In part b, a 90% confidence interval can be constructed to estimate the proportion of all U.S. adults.

Step-by-step explanation:

a. To answer part a of the question, we can perform a hypothesis test to determine if 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. We can set up the null and alternative hypotheses as follows:

Null hypothesis (H0): The proportion in 2018 is greater than or equal to the proportion in 2014.

Alternative hypothesis (Ha): The proportion in 2018 is less than the proportion in 2014.

Using a 0.05 significance level, we can perform a one-tailed Z-test to test the hypotheses.

b. To answer part b of the question, we can use the sample data to construct a 90% confidence interval for the proportion of all U.S. adults who planned to watch at least a 'fair amount' of the sporting event in 2018. This confidence interval will support the hypothesis test conclusion by providing an estimate of the range within which the true population proportion likely falls.

User Shd
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5 votes

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.

User Adnrw
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8.4k points

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