Question

“Streaming” of television programs is trending upward. According to The Harris Poll (August 26, 2013), over one-third of American’s qualify as “subscription streamers,” i.e., those who watch streamed TV programs through a subscription service such as Netflix, Hulu Plus, or Amazon Prime. The poll included 2,242 adult TV viewers, of which 785 are subscription streamers. On the basis of this result, can you conclude that the true fraction of adult TV viewers who are subscription streamers differs from one-third? Carry out the test using a Type I error rate of $\mathit{\alpha }\mathbf{=}\mathbf{.}\mathbf{10}$. Be sure to give the null and alternative hypotheses tested, test statistic value, rejection region or p-value, and conclusion.

Short answer

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.

Step-by-step solution

Given information

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

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

Setting up the hypotheses

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

$H_a:p\ne \frac{1}{3}$

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

 Step 3: Calculating test statistic value

The test statistic for testing these hypotheses is given as

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

Calculating the p-value

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

Conclusion using p-value

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.