Respuesta :
Answer:
Confidence level of 98%.
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].
Estimate:
The estimate [tex]\pi[/tex] is the mean of the two endpoints of the confidence interval. So
[tex]\pi = \frac{0.16+0.24}{2} = \frac{0.40}{2} = 0.20[/tex]
Survey of 545 television viewers
This means that [tex]n = 545[/tex]
What level of confidence (to the nearest percent, not a proportion) did the statistician use in constructing this interval
First we find z. The upper end is 0.24. So
[tex]0.20 \pm z\sqrt{\frac{0.2*0.8}{545}} = 0.24[/tex]
[tex]z\sqrt{\frac{0.2*0.8}{545}} = 0.04[/tex]
[tex]0.017z = 0.04[/tex]
[tex]z = \frac{0.04}{0.017}[/tex]
[tex]z = 2.33[/tex]
[tex]z = 2.33[/tex] has a pvalue of 0.99.
So
[tex]1 - \frac{\alpha}{2} = 0.99[/tex]
[tex]\frac{\alpha}{2} = 0.01[/tex]
[tex]\alpha = 0.02[/tex]
[tex]1 - \alpha = 1 - 0.02 = 0.98[/tex]
So a confidence level of 98%.
