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natita [175]
3 years ago
7

In a study conducted at a large long-term care facility, two data collectors examined 56 pressure ulcers on 40 different subject

s. The examinations were independently performed on the same day. A comparison of the results indicated that the data collectors scored 54 of the 56 (correlation coefficient 0.96) pressure ulcers identically using the Braden Scale for pressure ulcer assessment. What can be determined from this finding?
Mathematics
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

Interrater reliability between the two data collectors was high

Step-by-step explanation:

Explanation-

  • Interrater reliability is the consistency of observations between two or more observers.
  • An Agreement of 54 out of 56 scores would indicate a high level of interrater reliability.
  • The data collection method was appropriate for pressure ulcer risk assessment.
  • The risk assessments did not have to be done by both evaluators at the same time.
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Find the area of the top, which is a circle.

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Round to the nearest tenth: 84.8 cubic inches

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Express the ratio as a decimal. Round to the nearest hundredth.
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Step-by-step explanation:

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The data to the right represent the weights​ (in grams) of a random sample of 50 candies. Complete parts​ (a) through​ (f). 0.92
Inessa05 [86]

Answer:

0.069

Step-by-step explanation:

The given data set is

0.92, 0.87, 0.88, 0.82, 0.82, 0.87, 0.97, 0.86, 0.89, 0.84, 0.81, 0.88, 0.77, 0.86, 0.93, 0.84, 0.72, 0.82, 0.74, 0.83, 0.93, 0.75, 0.79, 0.91, 0.84, 0.91, 0.88, 0.88, 0.83, 0.78, 0.99, 0.81, 0.78, 0.75, 0.82, 0.76, 0.82, 0.87, 0.91, 0.77, 0.72, 0.94, 0.71, 0.73, 0.81, 0.81, 0.86, 0.93, 0.93, 0.82.

Formula for mean:

Mean=\frac{\sum x}{n}

Sum of all terms = 41.98

Mean of the data set is

Mean=\frac{41.98}{50}

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Formula for standard deviation for population:

\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}

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\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n-1}}

\sigma=\sqrt{\frac{0.231792}{50-1}}

\sigma=\sqrt{0.004730449}

\sigma=0.06877826

\sigma\approx 0.069

Therefore, the standard deviation of the data set is 0.069.

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3 years ago
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