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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.

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.

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).

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

a)\(6.2\)

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

c)no

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