Respuesta :
Answer:
The 90% confidence interval for the population mean is [tex][0.5 - \frac{0.1645}{\sqrt{n}}, 0.5 + \frac{0.1645}{\sqrt{n}}][/tex], in which n is the number of students surveyed.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.645\frac{0.1}{\sqrt{n}} = \frac{0.1645}{\sqrt{n}}[/tex]
The lower end of the interval is the sample mean subtracted by M, while the upper end is M added to the sample mean of 0.5. Thus, the confidence interval is of:
The 90% confidence interval for the population mean is [tex][0.5 - \frac{0.1645}{\sqrt{n}}, 0.5 + \frac{0.1645}{\sqrt{n}}][/tex], in which n is the number of students surveyed.
