Question

the article ‘’evaluation of a ventilation strategy to prevent barotrauma in patients at high risk for Acute Respiratory Distress Syndrome” (New Engl. J. of Med., \(1998:355 - 358)\) reported on an experiment in which \(120\) patients with similar clinical features were randomly divided into a control group and a treatment group, each consisting of \(60\) patients. The sample mean ICU stay (days) and sample standard deviation for the treatment group were \(19.9\) and\(39.1\), respectively, whereas these values for the control group were \(13.7\) and \(15.8\).

a) Calculate a point estimate for the difference between true average ICU stay for the treatment and control groups. Does this estimate suggest that there is a significant difference between true average stays under the two conditions?

b) Answer the question posed in part (a) by carrying out a formal test of hypotheses. Is the result different from what you conjectured in part (a)?

c)Does it appear that ICU stay for patients given the ventilation treatment is normally distributed? Explain your reasoning9. The article “Evaluation of a Ventilation Strategy to Prevent Barotrauma in Patients at High Risk for

d)Estimate true average length of stay for patients given the ventilation treatment in a way that conveys information about precision and reliability

Short answer

the solution is

a)\(6.2\)

b) do not reject null hypothesis (at any reasonable confidence level)

c)no

d) \((10,21.9).\)

Step-by-step solution

calculate the point estimate for the difference between true average

given

\(\begin{array}{*{20}{l}}{\bar x = 19.9;}&{m = 60;}&{{s_1} = 39.1}\\{\bar y = 13.7;}&{n = 60;}&{{s_2} = 15.8.}\end{array}\)

(a)

The difference in true average between the groups has a point estimate of

\(\bar x - \bar y = 19.9 - 13.7 = 6.2\)

This point estimate suggests that the true averages of the two groups may differ.

find statistically significant difference

the normal test would be testing hypotheses \({H_0}:{\mu _1} - {\mu _2} = 0\) versus \({H_a}:{\mu _1} - {\mu _2} \ne 0\)

when the null hypothesis is

\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)

The test statistic value

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

The \(P\) value is derived for the acceptable alternative hypothesis by calculating the adequate area under the standard normal curve. When both \(m > 40,n > 40\), this test can be employed.

This approach can be used because both \(m,n > 40\). The value of \(z\) is.

\(z = \frac{{19.9 - 13.7}}{{\sqrt {39.{1^2}/60 + 15.{8^2}.60} }} = \frac{{6.2}}{{5.44}} = 1.14\)

The \(P\) value can be calculated as two times the area under the standard normal curve to the right of \(\left| z \right|\) because the alternative hypothesis is two-sided. The \(P\) value can be calculated using the table in the appendix or software.

\(P = 2 \times P(Z > 1.14) = 2 \times 0.1271 = 0.2542.\)

Since

\(P = 0.2542 > \alpha \)

For any reasonable \(\alpha \)

Do not reject null hypothesis

There is no statistically significant difference.

explain reason

c)

The length of time that patients spend in the ICU does not appear to be distributed evenly. The two standard deviations from the mean make the normal distribution symmetric.

\(19.9 + 2 \times 39.1 = 98.1\)

and the data is all greater than \(0\) (days), with the majority of the data falling between \(0\) and \(98.1\), indicating a positively skewed distribution (not symmetric).

confidence interval

d)

The normalized random variable for big \(n\).

\(Z = \frac{{\bar X - \mu }}{{S/\sqrt n }}\)

has a normal distribution with a standard deviation of \(1\) and an expectation of \(0\). As a result, a

large-sample confidence interval for \(\mu \)

\(\bar x \pm {z_{\alpha /2}} \times \frac{s}{{\sqrt n }}\)

with a degree of confidence of around \(100(1 - \alpha )\% \) Regardless of population distribution, this is true.

Because \(n = 60\) is a sufficiently big number, this confidence interval might be used to estimate the genuine average length of stay. For genuine average, the large-sample confidence interval is

\(\begin{array}{l}\left( {\bar x - {z_{\alpha /2}} \times \frac{s}{{\sqrt n }},\bar x + {z_{\alpha /2}} \times \frac{s}{{\sqrt n }}} \right) = \left( {19.9 - 1.96 \times \frac{{39.1}}{{\sqrt {60} }},19.9 + 1.96 \times \frac{{39.1}}{{\sqrt {60} }}} \right)\\ = (10,21.9).\end{array}\)

The confidence level chosen here is

\(\alpha = 0.05\) and \({z_{\alpha /2}} = {z_{0.025}} = 1.96.\)

conclusion

a)\(6.2\)

b) do not reject null hypothesis (at any reasonable confidence level)

c)no

d) \((10,21.9).\)