Answers To Questions On Drinking, CAGE Findings, And Validity Testing

How many people drink?

How many people drink?

Number of people who drink is 91+67+51+25=234

How many people have clinically significant CAGE findings that should be addressed?

	Frequency Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	Percent
0	917	80%
1	91	8%
2	67	6%
3	51	4%
4	25	2%

 Number of people with clinically significant CAGE findings that should be addressed is 67+51+25=143

The test-retest reliability was found to range from 0.80 to 0.95. What is this type of reliability and what does this indicate?

Internal consistency; Different questions, same construct.

The value indicates a high internal consistency (reliability) of the variables and proves that this research instrument was reliable.

The CAGE has been tested by comparing it to other alcohol screening tests, the AUDIT (Alcohol Use Disorders Identification Test) and the SMAST (short Michigan Alcoholism Screening Instrument) and found to have correlations of .70. 

This instrument testing is an example of what form of validity testing?

This is an example of concurrent or convergent validity, a form of construct validity.

 What does this form of testing tell us about the instrument?

It suggests a moderate to strong correlation supporting the construct validity of the instrument(r = 0.7).

Is there a need to reverse score any of the items in this survey: If so, which ones?

No there is no need to reverse score for any of the items in this survey

Provide the mean sum score & SD, median sum score, mode, and Interquartile Range.

Statistics
sum_score
N	Valid	1163
Missing	0
Mean	18.47
Median	14.00
Mode	13.00
Std. Deviation	7.88
Percentiles	25	13.00
50	14.00
75	22.00
Interquartile range		9.00

Is the data normally skewed? Support your answer. If not normally, skewed be sure to interpret your findings? 

The data is not normally skewed but is right skewed (having a longer tail to the right).

Write a short paragraph presenting and interpreting the descriptives on engagement scores- you do not need to include every single descriptive measure, rather those that are most relevant to this data set so that the reader can understand the meaning of the findings.

The average sum engagement score was found to be 18.47 with a sum median score of 14 while the most frequent sum score was 13. The interquartile range was 9. With a mean of 18.47 (which is a lower value) it implies that on average there is a greater provider engagement. Similarly, with a sum score of 13 being the most frequent it means that majority of the respondents always feel a greater provider engagement.

You want to determine whether HCP scores vary by gender.

What is the test you will select? Give your rationale.

Since the data was observed to be non-normally distributed Kruskal–Wallis test   will be the most ideal test.

Is there a difference in HCP scores by gender? Give your rational, provide the data.

Descriptive Statistics
	N	Mean	Std. Deviation	Minimum	Maximum
hcp_sum_score	1163	18.4695	7.87716	13.00	52.00
subject gender	1161	1.70	.482	1	3

Ranks
	subject gender	N	Mean Rank
hcp_sum_score	female	365	532.44
male	784	603.98
transgender	12	556.96
Total	1161

Test Statisticsa,b
	hcp_sum_score
Chi-Square	12.601
df	2
Asymp. Sig.	.002
a. Kruskal Wallis Test
b. Grouping Variable: subject gender

There was a statistically significant difference between the HCP sum score by different gender types (H(2)=18.47, p=0.002), with  a mean rank of 532.44 for the female respondents, 603.98 for the male respondents and 556.96 for the transgender respondents. 

Which groups differ from each other? Give your rationale, provide the data and interpretation of the data.

Post-Hoc test showed that there was significant difference in the scores of the females and transgender individuals. However, there was no difference among the males and the females nor between the males and the transgender individuals.

Multiple Comparisons
Dependent Variable:   sum_score
LSD
(I) subject gender	(J) subject gender	Mean Difference (I-J)	Std. Error	Sig.	95% Confidence Interval
Lower Bound	Upper Bound
female	male	-1.40514*	.49669	.005	-2.3797	-.4306
transgender	-.07648	2.29966	.973	-4.5885	4.4355
male	female	1.40514*	.49669	.005	.4306	2.3797
transgender	1.32866	2.28002	.560	-3.1448	5.8021
transgender	female	.07648	2.29966	.973	-4.4355	4.5885
male	-1.32866	2.28002	.560	-5.8021	3.1448
*. The mean difference is significant at the 0.05 level.

How many people have clinically significant CAGE findings that should be addressed?

 There are 8 items that need to be reverse-coded. What are these 8 items? 

1a.satisfied with physical activity

2a.enjoyed living

2b.control of life

2c.satisfied with socially active

2d.pleased with health

8a.i could see my hcp whenever i needed to

8b.hcp involves me in decision making

8c.hcp cares about me

Reverse code the 8 items.

Done

Using the recoded items:

What is the total mean score for HAT?

Descriptive Statistics
	N	Minimum	Maximum	Mean	Std. Deviation
hat_sum_score	1163	33.00	145.00	99.1006	21.77465
Valid N (listwise)	1163

 Assume the data to be normally distributed. You want to determine if there is a difference in scores by gender.

What test will you use? Give your rationale.

Analysis of Variance (ANOVA) test; this is because the factors to be compared are more than two and also we are aware that the data is assumed to be normally distributed. The dependent variable is also continuous hence all the above reasons suffice on using ANOVA.  

Is there a difference in scores by gender? (respond in a complete sentence providing data).

ANOVA
hat_sum_score
	Sum of Squares	df	Mean Square	F	Sig.
Between Groups	21839.843	51	428.232	.897	.680
Within Groups	521599.268	1092	477.655
Total	543439.111	1143

Analysis of variance (ANOVA) showed that there was no significant main effect for treatment (gender), F(51, 1092) = 0.897, p = .680. Thus we concluded that there no significant difference in HAT scores by gender (p-value > 0.05).

Assume the data to be normally distributed. You want to determine if Quality of Life Scores differ by literacy level (high and low).

What test will you use? Give your rationale.

Independent t-test would be ideal for use; the test compares the means between two unrelated groups (high and low) on the same continuous, dependent variable (Quality of Life Scores).

Are the scores significantly different by literacy level? (Respond in a complete sentence providing data).

Group Statistics
	Literacy level	N	Mean	Std. Deviation	Std. Error Mean
hat_sum_score	Low	159	102.6289	21.22053	1.68290
High	520	101.5346	20.14166	.88327

Independent Samples Test
	Levene’s Test for Equality of Variances	t-test for Equality of Means
F	Sig.	t	df	Sig. (2-tailed)	Mean Difference	Std. Error Difference	95% Confidence Interval of the Difference
Lower	Upper
hat_sum_score	Equal variances assumed	.162	.687	.592	677	.554	1.09432	1.84856	-2.53529	4.72392
Equal variances not assumed			.576	251.234	.565	1.09432	1.90061	-2.64884	4.83747

No the scores are not significantly different by the literacy levels (p-value > 0.05).  An independent samples t-test was done to compare the mean HAT sum scores based on literacy levels (low or high) of the respondents. Results showed that respondents with low literacy levels (M = 102.62, SD = 21.22, N = 159) had no significant difference in scores as compared to those with higher literacy levels (M = 101.53, SD = 20.14, N = 520), t (677) = 0.592, p =0.554, two-tailed.  Essentially results showed that education literacy level of the respondents has no effect on the HAT scores.

Based on the hypotheses, would you use a 1- or 2-tailed significance test? State your rationale.

1-tailed significance test would be ideal to test the hypothesis; the reason is we are testing whether the intervention performs better

Identify the independent and dependent variables in the study…. For each identified variable, identify how they were measured.

Independent:

External cold and vibration stimulation

How it was measured:

External cold and vibration stimulation via Buzzy

Dependent:

Fear and anxiety levels;

How it was measured:

The pre-procedural and procedural fear and anxiety levels of the children were assessed via self-parental and observer reports. Pre-procedural anxiety was assessed using the Children’s Fear Scale, along with reports by the children, their parents, and an observer

Procedural pain was assessed using the Wong Baker Faces Scale and the visual analog scale self-reports of the children.

The test-retest reliability was found to range from 0.80 to 0.95. What is this type of reliability and what does this indicate?

What is the population of interest? Discuss the strengths and weaknesses of the sampling and assignment methods.

Children during peripheral intravenous (IV) cannulation

The study employed Prospective randomized controlled trial

The strengths of this method are;

• The method eliminates bias in treatment assignment
• The method also facilitates blinding (masking) of the identity of treatments from investigators, participants, and assessors
• The method permits the use of probability theory to express the likelihood that any difference in outcome between treatment groups merely indicates chance.

The strengths of this method are;

• Conflict of interest may endanger the study e.g. if for instance the nurse had some conflict of interest then it is possible that this may influence the results in a negative way
• Time and cost; the method requires a time commitment and cost as well which may not be available.

In Table 1, the authors are comparing the 2 groups for similarity in demographic characteristics.

Why is this important to do and what do the results tell us?

This was important in determining group equivalency

Why was chi square used for the variable sex and a t-test for the remaining characteristics?

Sex is  a nominal variable and can be analyzed using Chi-Square test. T-test was used for the other characteristics since they were continuous variables.

Based on the data in Table 1, were the groups similar or different?

The groups were similar

What is the null hypothesis for observer-reported procedural anxiety for the two groups? Was this null hypothesis accepted or rejected in this study? Provide a rationale for your answer.

The null hypothesis is: There is no difference in observer-reported procedural anxiety levels between the Buzzy intervention and the control groups for school-age children. The t = -6.745 for observer-reported procedural anxiety levels, p = 0.000, which is less than alpha of 0.05 set for this study. Since the study result was significant, the null hypothesis was rejected.

What is the t-test result for BMI? Is this result statistically significant? Provide a rationale for your answer? What does this result mean for the study?

The mean BMI for the buzzy group was 25.41 (SD = 6.74) while that of the control group was 26.94 (SD = 8.68); conditions t(88) = –1.309, p = 0.192. This result is not statistically significant (p-value > 0.05). The results means that buzzy has no influence on the BMI.

Assuming the t-tests presented in Table 2 and Table 3 are all the t tests performed to analyze the dependent variables’ data, calculate a Bonferroni procedure for this study.

The Bonferroni procedure is calculated by alpha divided by number of t-tests conducted on study variable’s data. Note the researchers do not always report all t-tests conducted, especially if they were not statistically significant. The t-tests conducted on demographic data are not of concern. Canbulat reported the results of four t-tests conducted to examine the differences between the intervention and control groups for the dependent variables procedural self-reported pain with WBFS, procedural self-reported pain with VAS, parent-reported anxiety levels, and observer-reported anxiety levels. The Bonferroni calculation for this study: 0.05 (alpha) divided by number of t-tests conducted = 0.05/4 = 0.0125. The new alpha set for the study is 0.0125.

Would the t-test for observer-reported procedural anxiety be significant based on the more stringent alpha calculated using the Bonferroni procedure in question 7? Provide a rationale.

No; Based on the Bonferoni results = 0.0125, the t = -6.745, p = 0.000 is still significant since it is less than 0.0125.

Reference

Dancey, C., Reidy, J., & Rowe, R. (2012). Statistics for the health sciences: A non-mathematical introduction. Thousand Oaks, CA: Sage.

American Psychological Association (APA). (2010). Publication manual of the American Psychological Association, 6^th Ed. Washington, DC: APA. 

Dhalla, S., & Kopec, J. A. (2007). The CAGE Questionnaire for Alcohol Misuse: A Review of Reliability and Validity Studies. Clinical & Investigative Medicine,30(1), 33.

doi:10.25011/cim.v30i1.447

 Canbulat, N., Ayhan, F., & Inal, S. (2015). Effectiveness of External Cold and Vibration for Procedural Pain Relief During Peripheral Intravenous Cannulation in Pediatric Patients. Pain Management Nursing,16(1), 33-39. doi:10.1016/j.pmn.2014.03.003