# Nursing Statistics

Nursing Statistics

Exercise 14

1. The APLS formula overestimates the weight as shown by the fact that the line with black triangle is consistently higher than the blue squares.
2. Y=0.502(9)+3.161=7.679kgs
3. Y=0.502(2) +3.161=4.165kgs
4. Slope:0.716 kg/month y-intercept:7.241
5. Weight in kgs= (0.716*36)+7.241=17.801kgs
6. First calculate the weight of child in months, which is 5*12 months/year=60months.y=bx+a=0.716 (60)+7.241=17.801kgs
7. This is an expected finding considering the motive behind linear regression to observe statistical relationships, in which the relationship between the variables is not perfect.
8. First calculate the child’s age in months which is 10*12months=120months.

Y=(0.331*age in months)-6.868=0.331(120)-6.868=39.72-6.868=32.852kgs

1. The sample size of 10081 was adequate for conducting simple linear regression. A sample this big can effectively be considered a representative of normal population distribution.
2. One potential clinical advantage is the ability to utilize multiple formulas to effectively predict the estimated child weights in a PCIU setting in emergent cases. As mentioned In the study, there are instances where it necessitates beginning critical therapy immediately without taking the patient’s weight .These scenarios requires a reliable method of computing an estimated weight to be used in order to initiate effective therapy. A benefit to the patient and also the attending providers to ensure a positive prognosis.

One potential problem with using the three novel formula is the implications of commencing therapy based on a predicted weight .Without immediate acquisition of the actual weight, reliant of the novel formulas result may lead to erroneous dosing

Exercise 19

1. The categories that were reported to be statistically significant were provider types, practice setting, percentage of adolescent patients, years in practice and practice region.
2. The level of measurement that is appropriate is comparing frequencies observed with frequencies expected. When the x2 values are calculated, you compare the critical values x2 distribution table.
3. X=29.68; p<.00 therefore this is statistically significantly. Nursing research uses alpha=0.05 and the p value of this is p<.00
4. The df=(R-1) (c-1). This table contains 3 R rows which are health care providers, mental health provider and other. There are also 2 columns which are RAAPS users and non RAAPS users, therefore (3-1) (2-1) =2
5. There is a statistically significant difference between the users and nonusers of RAAPS. When looking at the chi square value we see that it is 12.7652 with a p<.00. Finding a difference between staff that use RAAPS because of education, training and availability would be expected.
6. There will be no difference for RAAPS users and RAAPS non users related to provide age in years.
7. The null hypothesis should be accepted when you see that x2=4.00 with a p value of 0.14 this indicates there is no difference.
8. One clinical advantage would be that the assessment is quick in that it can be done in 5-7 minutes. This can benefit it a crisis situation. A challenge would be if the institution does not allow for its use.
9. There are 5 null hypothesis in the study that are statistically significant as the p values were less than 0.05. This indicates that the null hypothesis is to be rejected. They are provider type practice settings, % adolescent patients’ years in the practice and US practice region.
10. The x2 does not provide the location of the difference. It shows that a statistical difference does not exist. In order to find the location of the difference a post hoc analysis would need to be performed

Exercise 29

1. The frequency distribution was not statistically significant as the Shapiro-Wilk p value was 0.357.
2. A student’s age is not a predictor for completion of the RN to BSN program.
3. b represents the slope of the regression line. b=0.047
4. A represents the y intercept or the point at which the regression line intersects the y axis.a=11
5. The new regression equation ifs y=0.47(age)+11.763
6. The magnitude of the obtained R2 value would be a small one according to the description after the lesson.
7. Knowing the student’s enrollment age, explains the variance in months of RN to BSN program completion, which is 1.2%.
8. Using the example and the new regression equation, the correlation between actual y values and predicted y values is 0.108.
9. Performing a simple linear regression with student age at time of program enrollment did not predict in a significant way how long a student would take to complete the program.
10. The effect noted when using the calculated regression equation was small therefore I do not believe I would use it. If we were to look at the results of the new regression equation however, it is noted that the differences for the youngest was 4.156 months and the oldest was 2.84.(youngest predicted y=0.047(23)+11.763.Oldest y=0.047(51)+11.763).As an older student ,I might find this disheartening

Exercise 35

1 yes the variables indicate that the categories are mutually exclusive as well of exhaustive. There is the presence of dartum per participant that is entered into the contingency table and both the antibiotic use and no antibiotic that are categorical or nominal level data

1. The Chi-square value is 11.931
2. Yes the p value=0.001 which is less than alpha =0.05
3. The assumption that the null hypothesis is true brings the exact likelihood of obtaining the x2 to at least as extreme or as close to the one that was observed which is 0.1%

5.15/15*100%=100% of the antibiotic users tested positive for candidaria

1. The percentage of non-antibiotic users who tested positive for candiduria is calculated as 0/39=0%
2. The percentage of veterans with candiduria who had a history of antibiotic use is calculated as 15/15=1=1*100%=100%
3. The percentage of veterans with candiduria who had no history of antibiotic use is calculated as 0/ 15=0=0*100%=0%
4. A person x2 analysis indicated that antibiotic users had significant higher rates of candiduria than those who did not use antibiotics, x2=11.931, p=0.001(25.9%). This finding suggest that extended antibiotic use may be a risk factor for developing candiduria and further research is needed to investigate the development of candidura as direct effect of antibiotics.

10The sample was adequate to detect difference between the two groups because a significant difference was found p=0.001, which is smaller than alpha= 0.05.

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