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fredd [130]
2 years ago
7

A client has orders to receive 2 liters of IV fluid over a 24-hour period with ½ this amount to be infused in the first 10 hours

of treatment. How many milliliters per hour would the client receive during the first 10 hours of the infusion?
Mathematics
1 answer:
Tcecarenko [31]2 years ago
4 0

Answer:100 ml/hr

Step-by-step explanation:

client has orders to receive 2 liters of IV fluid over a 24-hour period

½ this amount to be infused in the first 10 hours of treatment = ½ ×2 lires

= 1 litre =1× 000 ml= 1000ml

If 1000 ml is received in 10hrs

x ml will be received in 1 hrs

x = 1× 1000/10

= 100ml per hr

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Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi
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Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

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(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

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The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.

The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

The sample size is, <em>n</em> = 7.

The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.

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Use a <em>t</em>-table t compute the <em>p</em>-value.

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The <em>p</em>-value is:

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The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

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