Respuesta :
Answer:
The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
Of the 801 participants in the survey, 457 indicated they subscribed to at least one video-streaming service.
This means that [tex]n = 801, \pi = \frac{457}{801} = 0.5705[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5705 - 1.96\sqrt{\frac{0.5705*0.4295}{801}} = 0.5362[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5705 + 1.96\sqrt{\frac{0.5705*0.4295}{801}} = 0.6048[/tex]
The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).
