A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is

approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of

the tranquilizer for each patient (in hours) was as follows: $2.7;2.8;3.0;2.3;2.3;2.2;2.8;2.1;and2.4.$

$$\begin{gathered} a.i.x̄=\_\_\_\_\_\_\_\_\_\_ \\ ii.s_x=\_\_\_\_\_\_\_\_\_\_ \\ iii.n=\_\_\_\_\_\_\_\_\_\_ \\ iv.n–1=\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

b. Define the random variable $X$in words.

c. Define the random variable $\overline{X}$in words.

d. Which distribution should you use for this problem? Explain your choice.

e. Construct a $95\%$confidence interval for the population mean length of time.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

f. What does it mean to be “$95\%$ confident” in this problem?

i. To find the mean and standard deviation, use the formulas below.

To open the setup editor, hit STAT followed by the number $5$and then Enter. We must enter each value and then press the arrow to calculate, resulting in the final output seen below.

In the mean time,$\overline{x}=2.511$

ii. The sample length of time's standard deviation, $s_x=0.3179$

iii. The number of patients who were used in study, $n=9$.

iv. If the total number of patients involved in the study is $n=9than,n-1=8$.

For the effective length of X, the time for the tranquillizer.

From a sample of$9$ patients, the mean time for the tranquillizer for the effective length is $\overline{X}$.

The parameter is $t_{n-1}$, and the distribution is t.

$t_{9-1}=t_{8.}$

1. Using the $T I-83$ calculator, calculate the confidence interval. The confidence interval's output,

$CI=(2.2667,2.7555)$

ii. The graph is follows below:

iii. The error bound is calculated from the formula,

$EBM=t_{n-1}\left(\frac{\alpha }{2}\right)\frac{s}{\sqrt{n}}$

$EBM=t_{9-1}\left(\frac{0.05}{2}\right)\frac{0.31798}{\sqrt{9}}$

$EBM=0.2444$

If a group of nine patients were sampled, $90\%$of the sample would represent the real population mean length of time.