Question

Suppose you are a researcher in a hospital. You are experimenting with a new tranquilizer. You collect data from a random sample of 9 patients. The period of effectiveness of the tranquilizer for each patient (in hours) is as follows:

2.8

2.9

2.7

2

2.5

2.7

2

2.2

2.7

a. What is a point estimate for the population mean length of time. (Round answer to 4 decimal places)

b. Which distribution should you use for this problem? normal distribution or t-distribution

c. Why?

d. What must be true in order to construct a confidence interval in this situation? (Choose 1, 2, 3, or 4) 1. The population must be approximately normal 2. The sample size must be greater than 30 3. The population standard deviation must be known 4. The population mean must be known

e. Construct a 99% confidence interval for the population mean length of time. Enter your answer as an open-interval (i.e., parentheses) Round upper and lower bounds to two decimal places

f. Interpret the confidence interval in a complete sentence. Make sure you include units.

g. What does it mean to be "99% confident" in this problem? Use the definition of confidence level. (Choose 1, 2, or 3). 1. There is a 99% chance that the confidence interval contains the population mean 2. The confidence interval contains 99% of all samples 3. 99% of all simple random samples of size 9 from this population will result in confidence intervals that contain the population mean

h. Suppose that the company releases a statement that the mean time for all patients is 2 hours. Is it possible ? Yes or no Is it likely? Yes or no

i. Use the results above and make an argument in favor or against the company's statement. Structure your essay as follows:

1. Describe the population and parameter for this situation.
2. Describe the sample and statistic for this situation.
3. Give a brief explanation of what a confidence interval is.
4. Explain what type of confidence interval you can make in this situation and why.
5. Interpret the confidence interval for this situation.
6. Restate the company's claim and whether you agree with it or not.
7. Use the confidence interval to estimate the likelihood of the company's claim being true.
8. Suggest what the company should do.

Answer 1

Solution

Back-up Theory

Let X = period (in hours) of effectiveness of the tranquilizer

100(1 - α) % Confidence Interval for the population mean μ, when σ is not known is:

Xbar ± MoE………………………………………………………………………………….. (1)

where

MoE = (t_{n- 1, α /2})s/√n

with

Xbar = sample mean,

t_{n – 1, α /2} = upper (α/2)% point of t-distribution with (n - 1) degrees of freedom,

s = sample standard deviation and

n = sample size.

CENTRAL LIMIT THEOREM ……………………………………………………………. (2)

Let {X₁, X₂, …, X_n} be a sequence of n independent and identically distributed (i.i.d) random variables drawn from a distribution [i.e., {x₁, x_2, …, x_n} is a random sample of size n] of expected value given by µ and finite variance given by σ². Then, as n gets larger, the distribution of Z = {√n(Xbar − µ)/σ}, approximates the normal distribution with mean 0 and variance 1 (i.e., Standard Normal Distribution)

i.e., sample average from any distribution follows Normal Distribution with mean µ and variance σ²/n

Now to work out the solution,

Details of calculation follow at the end.

Part (a)

Point estimate for the population mean length of time

= Sample mean, Xbar

= 2.5000 hours Answer 1

Part (b)

Distribution one should use for this problem is: t-distribution Answer 2

Part (c)

By Central Limit Theorem, [vide (2)], Xbar ~ Normal and since population standard deviation is not known, it must be estimated by sample standard deviation. Hence, the distribution is t. Answer 3

Part (d)

By CLT [vide (2)], Xbar ~ Normal so far as n ≥ 30. Hence, Option 2 Answer 4

Part (e)

Vide (1), 99% confidence interval for the population mean length of time is:

(2.11, 2.89) hours. Answer 5 [Details of calculation follow at the end.]

Part (f)

99% of the times, the population mean would be between 2.11 and 2.89. Answer 6

Part (g)

Option 1 [see interpretation under (f)] Answer 7

Part (h)

Company’s statement that the mean time for all patients is 2 hours is not possible because the the confidence interval does not hold 2. Answer 8

Is it likely? Yes. There is 0.005 chance that it could be 2. Answer 9

DONE

Calculation Details

Data

i	xi
1	2.8
2	2.9
3	2.7
4	2
5	2.5
6	2.7
7	2
8	2.2
9	2.7

Summary of Calculations

n	9
Xbar	2.5000
s	0.34641
√n	3
α	0.01
n - 1	8
tα/2	3.355387
(s/√n)(tα/2)	0.387447
Lower Bound	2.112553
Upper Bound	2.887447

DONE