Question

Question 17 options:

Answer 1

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:

$\bar x=\frac{1}{n}\sum X=\frac{54.3 }{8}=6.788\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}=\sqrt{\frac{1}{8-1}\times 11.8288}=1.299$

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:

$t_{\alpha/2, (n-1)}=t_{0.04, 7}=2.046$

*Use a t-table.

Compute the 92% confidence interval for the true mean time a student sleeps per night as follows:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\frac{s}{\sqrt{n}}$

     $=6.788\pm 2.046\times\frac{1.299}{\sqrt{8}}\\\\=6.788\pm 0.9392\\\\=(-0.2504, 1.6080)\\\\\approx (-0.25, 1.61)$

We have 92 % confidence that the true mean time a student sleeps per night is between (-0.25, 1.61).