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As per the Harris Poll, out of 2242 adult TV viewers surveyed, 785 are subscription streamers.

That is

The size of the sample$n=2242$

The sample proportion is

$$\begin{gathered} \hat{p}=\frac{785}{2242} \\ =0.350 \end{gathered}$$

We have to test whether the true fraction of adult TV viewers who are subscription streamers differs from one-third.

As per the scenario, the null and alternative hypothesis is

$H_0:p=\frac{1}{3}=0.3333$

That is, the true fraction of adult TV viewers who are subscription streamers does not differ from one-third.

And

$H_a:p鈮�\frac{1}{3}$

That is, the true fraction of adult TV viewers who are subscription streamers differs from one-third.

The test statistic for testing these hypotheses is given as

$$\begin{gathered} Z=\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} \\ =\frac{0.350-0.3333}{\sqrt{\frac{0.3333\left(1-0.3333\right)}{2242}}} \\ =\frac{0.0167}{\sqrt{0.000099113}} \\ =1.68 \end{gathered}$$

We have $Z=1.68$, and the test is two-tailed (as an alternative hypothesis is two-tailed)

Therefore p-value in this scenario is

localid="1668671294086" $$\begin{gathered} p-value=2脳P\left(Z>Z_0\right) \\ =2脳P\left(Z>1.68\right) \\ =2脳0.0465.....\left[usingstandardnormaltable\right] \\ =0.0930 \end{gathered}$$

We can see that

$p-value=0.0930<0.10$

That is, the obtained p-value is less than the significance level.

Hence, we reject the null hypothesis.

Conclusion:

At a 10% significance level, we have sufficient evidence to conclude that the true fraction of adult TV viewers who are subscription streamers differs from one-third.