A pharmaceutical company makes tranquilizers. It is assumed that the distributio

Question

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 9 patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.6; 2.9; 3.1; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

(i) Round your answer to two decimal places.

x =

(ii) Round your answer to two decimal places.

sx =

(iii) Enter an exact number as an integer, fraction, or decimal.

n =

(iii) Enter an exact number as an integer, fraction, or decimal.

n 1 =

2. Which distribution should you use for this problem? (Enter your answer in the form z or tdf where df is the degrees of freedom.)

3. Construct a 95% confidence interval for the population mean length of time.

(i) State the confidence interval. (Round your answers to two decimal places.)

(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to two decimal places.)

Explanation / Answer

1.
sample mean, x =2.5222 ~ 2.52
standard deviation, s =0.3456 ~
number (n)=9, n-1 = 8
2.
it is student t distribution
the value of |t | with n-1 = 8 d.f
3.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=2.5222
Standard deviation( sd )=0.3456
Sample Size(n)=9
Confidence Interval = [ 2.5222 ± t a/2 ( 0.3456/ Sqrt ( 9) ) ]
= [ 2.5222 - 2.306 * (0.115) , 2.5222 + 2.306 * (0.115) ]
= [ 2.257,2.788 ]

Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size
Standard deviation( sd )=0.3456
Sample Size(n)=9
Standard Error = ( 0.3456/ Sqrt ( 9) )
= 0.115

Interpretations:
1) We are 95% sure that the interval [2.257 , 2.788 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean  

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.