Respuesta :
Answer:
We have 92 % confidence that the true mean time a student sleeps per night is between (-0.25, 1.61).
Step-by-step explanation:
Compute the sample mean and sample standard deviation as follows:
[tex]\bar x=\frac{1}{n}\sum X=\frac{54.3 }{8}=6.788\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 11.8288}=1.299[/tex]
Since, the sample standard deviation is computed the t-statistics will be used for the confidence interval.
The critical value of t for (n - 1) degrees of freedom and 92% confidence level is:
[tex]t_{\alpha/2, (n-1)}=t_{0.04, 7}=2.046[/tex]
*Use a t-table.
Compute the 92% confidence interval for the true mean time a student sleeps per night as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\frac{s}{\sqrt{n}}[/tex]
[tex]=6.788\pm 2.046\times\frac{1.299}{\sqrt{8}}\\\\=6.788\pm 0.9392\\\\=(-0.2504, 1.6080)\\\\\approx (-0.25, 1.61)[/tex]
We have 92 % confidence that the true mean time a student sleeps per night is between (-0.25, 1.61).
