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3 years ago

5

A solution consisting of 182 mg of dopamine in 26 mL of solution is administered at a rate of 10 mL/hr. Complete parts (a) and

(b) below
What is the flow rate in mg of dopamine per hour?
. If a patient is prescribed to receive 140 mg of dopamine, how long should the infusion last?

1 answer:

Cloud [144]3 years ago

8 0

Answer:

<u>The correct answer is A. 70 mg and B. 2 hours.</u>

Step-by-step explanation:

A. Let's calculate the flow rate of dopamine per hour:

Flow rate of dopamine per hour = (182 * 10)/26

Flow rate of dopamine per hour = 1,820/26

<u>Flow rate of dopamine per hour = 70 mg</u>

B. Let's calculate how much the solution should last for a patient that is prescribed to receive 140 mg of dopamine

Prescription = 140 mg

Flow rate of dopamine per hour = 70 mg

Duration = Prescription/Flow rate of dopamine per hour

Duration = 140/70

<u>Duration = 2 hours</u>

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Answer:

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And we can find the p value using the following excel code:

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

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Freshman 23 21 43 21 108

Sophomore 34 23 14 26 97

Junior 25 30 23 26 104

Senior 27 33 17 19 96

Total 109 107 97 92 405

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two random variables

H1: There is dependence between the two variables

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{109*108}{405}=29.07$

$E_{2} =\frac{107*108}{405}=28.53$

$E_{3} =\frac{97*108}{405}=25.87$

$E_{4} =\frac{92*108}{405}=24.53$

$E_{5} =\frac{109*97}{405}=26.11$

$E_{6} =\frac{107*97}{405}=25.63$

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Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(4-1)(4-1)=9$

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And we can find the p value using the following excel code:

"=1-CHISQ.DIST(27.356,9,TRUE)"

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