5.2004. In 2014 (population 1), 21 children were sampled and in 2004 (population 2) another sample of 21 wastaken. Results for many test are shown in the table. Do the data indicate that the mean hospital stay for childrenhas decreased? Test with a significance level of 5%.The mean length hospital stay for children treated for pneumonia was compared between 2014 and20142004Sample meanSample standard deviationSample size3.74.31.51.22121

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Description: 5.2004. In 2014 (population 1), 21 children were sampled and in 2004 (population 2) another sample of 21 wastaken. Results for many test are shown in the table. Do the data indicate that the mean hospital stay for childrenhas decreased? Test with a

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MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

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5.
2004. In 2014 (population 1), 21 children were sampled and in 2004 (population 2) another sample of 21 was
taken. Results for many test are shown in the table. Do the data indicate that the mean hospital stay for children
has decreased? Test with a significance level of 5%.
The mean length hospital stay for children treated for pneumonia was compared between 2014 and
2014
2004
Sample mean
Sample standard deviation
Sample size
3.7
4.3
1.5
1.2
21
21

Expert Solution

Step 1

Given
The data of mean length hospital stay for children for pneumonia in 2004 and 2014 is shown below
$\begin{matrix} 2014 & 2004 \\ S a m p l e m e a n & 3.7 & 4.3 \\ S a m p l e S \tan d a r d d e v i a t i o n & 1.5 & 1.2 \\ S a m p l e s i z e & 21 & 21 \end{matrix}$

The level of significance is 5% or 0.05.

Let $μ_{1} μ_{2}$ be the mean length hospital stay for children in 2014 and 2004.

The null and alternate hypothesis are shown below
$H_{0} : μ_{1} = μ_{2} H_{a} : μ_{1} < μ_{2}$

The test statistic is calculated as shown below
$t = \frac{\overset{__}{x_{1}} - \overset{__}{x_{2}}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}} t = \frac{3.7 - 4.3}{\sqrt{\frac{1 . 5^{2}}{21} + \frac{1 . 2^{2}}{21}}} = - 1.43$

The degree of freedom is calculated using the below mentioned formula
$d f = \frac{{(\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}})}^{2}}{\frac{{(\frac{s_{1}^{2}}{n_{1}})}^{2}}{n_{1} - 1} + \frac{{(\frac{s_{2}^{2}}{n_{2}})}^{2}}{n_{2} - 1}} = \frac{{(\frac{1 . 5^{2}}{21} + \frac{1 . 2^{2}}{21})}^{2}}{\frac{{(\frac{1 . 5^{2}}{21})}^{2}}{21 - 1} + \frac{{(\frac{1 . 2^{2}}{21})}^{2}}{21 - 1}} = 38.16$

Thus we use degree of freedom as 38

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