Respuesta :
Answer:
70 students need to be 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 - 0.95}{2} = 0.025[/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 - 0.025 = 0.975[/tex], so Z = 1.96.
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.
Assuming that the standard deviation of the above sample is a reasonable estimate of the standard deviation of sleep time at your university.
This means that [tex]\sigma = 2.12[/tex]
How many students do you need to survey to estimate the mean sleep time of students at your university with 95% confidence and a margin of error of 0.5 hours?
n students need to be surveyed, and n is found when M = 0.5. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.96\frac{2.12}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 1.96*2.12[/tex]
Multiplying by 2 on both sides:
[tex]\sqrt{n} = 1.96*4.24[/tex]
[tex](\sqrt{n})^2 = (1.96*4.24)^2[/tex]
[tex]n = 69.1[/tex]
Rounding up
70 students need to be surveyed.
