Surgery patients: proportiona of patients with low pH

fastfred

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Can anyone help me solve this problem?

Table of pH measurements:___________________________
Morning patients:. . .. .n<sub>1</sub> = 50. . .(x-bar)<sub>1</sub> = 3.94. . .s<sub>1</sub> = 2.51
Afternoon patients:. . .n<sub>2</sub> = 49. . .(x-bar)<sub>2</sub> = 2.93. . .s<sub>2</sub> = 2.39

Surgery patients are fasted, that is, not allowed food for some period of time, before they are anesthetized. This is done to reduce the risk of patients inhaling their stomach contents. If this happens, they may develop a severe pneumonia-like condition called Mendelson's syndrome, which often results in death. The risk of getting Mendelson's syndrome is increased if the inhaled stomach contents are particularly acidic (that is, if they have a particularly low pH).

A study at [name deleted] Hospital led by Dr. Alan Merry looked at the effect of length of fast on stomach pH levels. Patients operated on in the morning have shorter fasts than those operated on in the afternoon. An interesting question is, "Does the additional fasting change average stomach pH levels?" The data from the study are given in the table above.

a) Is there any evidence of a difference between mean pH levels for morning and afternoon patients? Test at a = 0.05.

It is widely believed that the risk of Mendelson's syndrome is increased if the inhaled stomach contents have a pH lower than 2.5. In this study, 21 of the 50 morning patients and 31 of the 49 afternoon patients had a stomach pH below 2.5.

b) Is the proportion of afternoon patients with a pH below 2.5 significantly larger than the proportion of morning patients with a pH below 2.5? Test at a = 0.05.
 
What are the null and alternative hypotheses?.

\(\displaystyle H_{0}:{\mu}_{1}={\mu}_{2}\)

\(\displaystyle H_{a}:{\mu}_{1}\neq{\mu}_{2}\)


Because the test is two-tailed and you're using a 0.05 level of significance, the critical values are -1.96 and 1.96.

The rejection regions are \(\displaystyle z<-1.96\;\ and\;\ z>1.96\)

\(\displaystyle \L\\{\sigma}=\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n^{2}}}\)

Is z in the rejection region?. If so, reject. If it's not, fail to reject.




\(\displaystyle \L\\z=\frac{(\overline{x}_{1}-\overline{x}_{2})}{\sigma}\)
 
thanks.

SD=0.47

so do not reject

z=2.14

how do i interpret this?

is this correct

No, the proportion of afternoon patients with a pH below 2.5 is not significantly larger than the proportion of morning patients with a pH below 2.5
 
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