Post RN BSN Biostatistics – Unit V Making Inference from Data Solved Answers

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Posted by POST RN BSN Resources · Original syllabus post: May 4, 2017 · Category: Post RN BSN Nursing Course Syllabus

Post RN BSN Nursing – Biostatistics, Course No. 717 – Part 5 of 6. Expanded exam-style model answers for Unit V of the LUMHS Post RN BSN biostatistics syllabus: sampling and standard error, hypothesis testing, p value, Z-test, confidence interval, t-tests and chi-square, with worked examples. Use them as a study guide and write in your own words in the exam hall.

Please note: The syllabus lists topics only. These are exam-style questions and model answers prepared from those topics; they are not an official LUMHS paper or answer key. Worked examples use made-up numbers for practice.

📚 Course syllabus: Biostatistics (Course No. 717), LUMHS Post RN BSN – read the full course syllabus here.

Biostatistics Unit V Inference Solved: Z-test, t-test, Chi-Square Use as a revision guide and verify with your course notes.

Unit V: Making Inference from Data

Course syllabus · Model exam answers

B.Sc NURSING (POST-RN), SECOND YEAR, SEMESTER-III

BIOSTATISTICS – COURSE NO. 717

UNIT V: MAKING INFERENCE FROM DATA

Credit hours: 3Source: course syllabus

Evaluation: Assignment 10%, Internal 10%, Final 80%

Syllabus topics: Making inference from data; Z-test; confidence interval; p value; t-test; chi-square.

Note: the syllabus lists “conflict interval”, which is understood here as confidence interval. All data in the examples are made up for practice. Real studies use software; hand calculations are shown only to explain the logic.

Q-1. What is statistical inference? Distinguish between a population and a sample, and a parameter and a statistic.

Statistical inference is the process of drawing conclusions about a population from a sample. It has two parts: estimation (point estimate and confidence interval) and hypothesis testing (tests of significance).

Term	Meaning	Symbol / example
Population	The whole group of interest	All diabetic patients in a district
Sample	A part of the population actually studied	200 diabetic patients selected
Parameter	A numerical value describing a population	Population mean μ, proportion π, SD σ
Statistic	A numerical value describing a sample	Sample mean x̄, proportion p, SD s

Because different samples give different statistics, a sample result is only an estimate. Inference measures how far it can be trusted. A sample that is not random or representative gives misleading inference however good the test.

Q-2. What is sampling variation, sampling distribution and standard error?

Sampling variation: different samples from the same population give slightly different means. The sampling distribution of the mean is the distribution of means from all possible samples of the same size; it is approximately normal for reasonably large samples (central limit theorem) and has the population mean at its centre.

Standard error of the mean (SE) = s ÷ √n SE of a proportion = √[ p(1 − p) ÷ n ]

SD describes the spread of individual values; SE describes the spread of sample means (precision of the estimate).
SE becomes smaller as the sample size increases; quadrupling the sample halves the SE.

Example (made up): SD of systolic BP = 15 mmHg. With n = 36, SE = 15 ÷ 6 = 2.5 mmHg; with n = 144, SE = 15 ÷ 12 = 1.25 mmHg.

Q-3. What is a hypothesis? Explain null and alternative hypotheses and the steps of testing a hypothesis.

A hypothesis is a statement about a population that can be tested with data.

Null hypothesis (H₀): there is no difference or no association (e.g. the new dressing and the old dressing heal wounds equally fast). It is the statement tested.
Alternative hypothesis (H₁ or H_a): there is a difference or association. It may be two-tailed (the means differ) or one-tailed (the new dressing heals faster).

Steps of a test of significance

State H₀ and H₁.
Choose the level of significance (α), usually 0.05.
Select the suitable test (Z, t, chi-square) according to the data.
Calculate the test statistic from the sample.
Find the p value or compare with the critical value.
Decide: reject or do not reject H₀.
State the conclusion in plain words.

We never “prove” H₀ true; we either have enough evidence against it or we do not.

Q-4. Explain Type I and Type II errors, level of significance and power.

	H₀ actually true	H₀ actually false
Reject H₀	Type I error (false positive), probability = α	Correct (probability = power)
Do not reject H₀	Correct	Type II error (false negative), probability = β

Level of significance (α): the maximum risk of Type I error we accept, commonly 5% (0.05).
Power = 1 − β: the chance of detecting a real difference. Larger samples, bigger true differences and less variability increase power.
Lowering α reduces Type I errors but increases Type II errors for the same sample size.

Example: concluding that a new training reduces infections when it really does not = Type I error; concluding it does not work when it really does = Type II error.

Q-5. Define p value. How is it interpreted? What are its common misunderstandings?

The p value is the probability of getting a result at least as extreme as the one observed if the null hypothesis were true.

If p ≤ α (usually 0.05): the result is statistically significant; reject H₀.
If p > α: not significant; there is not enough evidence to reject H₀.

Rejection regions of a two-tailed Z-test: values beyond ±1.96 have a combined probability of 5%.

Common misunderstandings

p is not the probability that H₀ is true.
p = 0.03 does not mean the result is large or clinically important. A very large study can give a tiny p for a trivial difference.
p > 0.05 does not prove there is no difference; the sample may have been too small.
The 0.05 cut-off is a convention, not a law; report the exact p value (e.g. p = 0.016) and the confidence interval.

Q-6. Explain the Z-test. When is it used? Solve an example.

The Z-test is used to compare a sample mean with a known population mean (or two large sample means or proportions) when the sample is large (usually n ≥ 30) or the population SD is known and the data are approximately normal.

One-sample Z = (x̄ − μ₀) ÷ (σ ÷ √n)

Example (made up): the mean systolic BP of adults in a district is believed to be 120 mmHg (SD 15). A nurse measures 36 randomly selected adults of a community and finds a mean of 126 mmHg. Is the community different?

H₀: μ = 120; H₁: μ ≠ 120 (two-tailed); α = 0.05.
SE = 15 ÷ √36 = 15 ÷ 6 = 2.5.
Z = (126 − 120) ÷ 2.5 = 2.4.
Critical value at 5% (two-tailed) = ±1.96. Since 2.4 > 1.96, H₀ is rejected. The exact p value = 0.016.
Conclusion: the community mean BP is significantly different from 120 mmHg (p = 0.016).

Z critical values	Two-tailed	One-tailed
α = 0.05	±1.96	1.645
α = 0.01	±2.58	2.33

Z-test for two proportions (idea): Z = (p₁ − p₂) ÷ SE of the difference; it is used, for example, to compare vaccination coverage in two areas. For a 2 × 2 table the chi-square test (below) gives the same conclusion.

Q-7. What is a confidence interval (CI)? How is it interpreted? Calculate with examples.

A confidence interval is a range of values, calculated from a sample, that is likely to contain the true population value. A 95% CI means that if the study were repeated many times, about 95% of such intervals would contain the true value.

A confidence interval is the sample estimate ± the margin of error.

95% CI for a mean = x̄ ± (critical value × SE) 95% CI for a proportion = p ± 1.96 × √[p(1 − p) ÷ n]

Example 1 – mean (made up): haemoglobin of the 30 women in Unit III: mean = 11.0 g/dL, SD = 1.29. SE = 1.29 ÷ √30 = 0.236. For a small sample the t critical value (df = 29) is 2.045, so CI = 11.0 ± 2.045 × 0.236 = 11.0 ± 0.48 = 10.52 to 11.48 g/dL.

Example 2 – proportion (made up): 48 of 160 patients (p = 0.30) had a respiratory infection. SE = √(0.30 × 0.70 ÷ 160) = 0.0362. 95% CI = 0.30 ± 1.96 × 0.0362 = 0.30 ± 0.071 = 22.9% to 37.1%.

Interpretation

A narrow CI means a precise estimate; a wide CI means an imprecise estimate (usually a small sample).
Higher confidence (99%) gives a wider interval; a larger sample gives a narrower interval.
If the CI for a difference includes 0 (or for a ratio includes 1), the result is not significant at the 5% level.
CIs show both significance and the size of the effect, so they are more informative than a p value alone.

Q-8. Explain the t-test. Why is it used? State its types and assumptions.

The t-test (Student’s t-test) is used to compare means when the sample is small (n < 30) or the population SD is unknown and the data are approximately normal. The t distribution is similar to the normal curve but with heavier tails; its shape depends on the degrees of freedom (df), and approaches the normal curve as df increases.

Type	Used to compare	Degrees of freedom
One-sample t-test	A sample mean with a known or standard value	n − 1
Independent (unpaired) t-test	Means of two separate groups (e.g. intervention vs control)	n₁ + n₂ − 2
Paired t-test	Two measurements on the same people (before and after)	n pairs − 1

Assumptions

Numerical (interval or ratio) data.
Approximately normal distribution (or reasonably large sample).
Observations are independent (except pairs in the paired test).
Similar spread in both groups for the independent test.

If the data are clearly skewed or ordinal, non-parametric tests (e.g. Mann–Whitney U, Wilcoxon signed-rank) are used instead.

Q-9. Solve an example of the paired t-test.

Example (made up): systolic BP (mmHg) of 8 patients before and after a 4-week lifestyle programme.

Patient	Before	After	Difference (before − after)
1	150	142	8
2	142	138	4
3	160	150	10
4	155	150	5
5	148	144	4
6	138	136	2
7	152	146	6
8	146	140	6
Mean			5.625

H₀: mean difference = 0; H₁: mean difference ≠ 0; α = 0.05.
Mean difference = 45 ÷ 8 = 5.625; SD of the differences = 2.504.
SE = 2.504 ÷ √8 = 0.885.
t = 5.625 ÷ 0.885 = 6.35, df = 8 − 1 = 7.
Critical t (df = 7, two-tailed, 5%) = 2.365. Since 6.35 > 2.365, H₀ is rejected (exact p = 0.0004).
Conclusion: the programme was associated with a significant fall in systolic BP (mean fall 5.6 mmHg, 95% CI 3.5 to 7.7).

Caution: without a control group, a before–after fall cannot prove that the programme was the cause.

Q-10. Solve an example of the independent (unpaired) t-test.

Example (made up): pain scores (0–10) after two different positioning techniques, six patients in each group.

Group	Scores	n	Mean	SD
A (usual position)	6, 7, 5, 8, 6, 7	6	6.5	1.049
B (new position)	4, 5, 3, 6, 4, 5	6	4.5	1.049

H₀: the two population means are equal; H₁: they differ; α = 0.05.
Pooled SD = 1.049 (the two SDs are equal here, so the pooled SD equals them).
SE of the difference = 1.049 × √(1÷6 + 1÷6) = 0.606.
t = (6.5 − 4.5) ÷ 0.606 = 3.30, df = 6 + 6 − 2 = 10.
Critical t (df = 10, 5%, two-tailed) = 2.228. Since 3.30 > 2.228, H₀ is rejected (exact p = 0.008).
Conclusion: mean pain is significantly lower with the new position (difference 2.0 points). Pain scores are ordinal, so in real research a rank-based test might be preferred; this is shown only to illustrate the method.

Q-11. Explain the chi-square test. State its uses, assumptions and formula.

The chi-square (χ²) test is used for categorical (nominal or ordinal) data presented as counts. It tests (1) the association between two categorical variables (test of independence) and (2) how well observed counts fit expected counts (goodness of fit).

χ² = Σ [ (O − E)² ÷ E ] Expected count = (row total × column total) ÷ grand total df = (rows − 1) × (columns − 1)

Assumptions

Data are counts (frequencies), not percentages or means.
Observations are independent and categories do not overlap.
Expected count should be at least 5 in most cells (≥ 80%); otherwise combine categories or use Fisher’s exact test for 2 × 2 tables.

Q-12. Solve an example of the chi-square test of association.

Example (made up): hospital-acquired infection among patients cared for by nurses who did or did not receive a hand-hygiene training programme.

	Infection	No infection	Total
Trained nurses	12	88	100
Untrained nurses	28	72	100
Total	40	160	200

H₀: infection is not associated with training; H₁: it is associated; α = 0.05.
Expected counts: infection in each group = 100 × 40 ÷ 200 = 20; no infection = 100 × 160 ÷ 200 = 80.

Cell	O	E	(O − E)² ÷ E
Trained, infection	12	20	64 ÷ 20 = 3.2
Trained, no infection	88	80	64 ÷ 80 = 0.8
Untrained, infection	28	20	64 ÷ 20 = 3.2
Untrained, no infection	72	80	64 ÷ 80 = 0.8
χ² total			8.0

df = (2 − 1)(2 − 1) = 1. Critical χ² (df = 1, 5%) = 3.84; (1%) = 6.63.
Since 8.0 > 3.84 (and > 6.63), H₀ is rejected; p = 0.005.
Conclusion: infection is significantly associated with training: 12% in trained vs 28% in untrained nurses’ patients.

A significant association does not by itself prove cause (other differences between wards may exist).

Q-13. Which test should be used? (Selection of test.)

Question	Data type	Suitable test
One sample mean vs known value, large n or known SD	Numerical	One-sample Z-test
One sample mean vs known value, small n	Numerical	One-sample t-test
Two separate groups, compare means	Numerical, approximately normal	Independent t-test (Z-test if large n)
Same people, before and after	Numerical	Paired t-test
Two or more groups, compare proportions / association	Categorical counts	Chi-square test (Fisher’s exact if small expected counts)
Skewed or ordinal data	Ordinal / skewed numerical	Non-parametric tests (Mann–Whitney, Wilcoxon)
Three or more group means	Numerical	ANOVA (beyond this syllabus)
Relationship between two numerical variables	Numerical	Correlation / regression (beyond this syllabus)

Q-14. Distinguish between statistical significance and clinical significance. What are the common errors in inference?

Statistical significance tells whether a result is unlikely due to chance; clinical significance asks whether the difference is large enough to matter for patients. A fall of 1 mmHg may be “significant” in a huge sample but of no clinical value.
Choosing a test that does not fit the data type.
Drawing conclusions from a small or biased (non-random) sample.
Testing many hypotheses and reporting only the significant ones (multiple testing).
Confusing association with causation.
Reporting only p values without the size of the effect and the confidence interval.

Q-15. Solve these problems (practice exercise with made-up data).

Problem 1. A sample of 25 patients has mean haemoglobin 10.4 g/dL and SD 1.5. Find the 95% CI (t for df 24 = 2.064).

Answer: SE = 1.5 ÷ 5 = 0.3; margin = 2.064 × 0.3 = 0.62; CI = 10.4 ± 0.62 = 9.78 to 11.02 g/dL.

Problem 2. The calculated Z is 1.5 in a two-tailed test at 5%. Decision?

Answer: 1.5 < 1.96, so H₀ is not rejected (not significant).

Problem 3. In a chi-square test of a 3 × 2 table, what is the df?

Answer: (3 − 1)(2 − 1) = 2.

Problem 4. A trial reports a mean difference of 3.0 mmHg, 95% CI −1.0 to 7.0, p = 0.14. Interpret.

Answer: the CI includes 0 and p > 0.05, so the difference is not statistically significant. The interval is wide, so the study may have been too small to be sure; a real difference is not ruled out.

Problem 5. Which test compares blood pressure before and after drug in the same 12 patients?

Answer: paired t-test.

Q-16. What is the importance of inferential statistics in nursing research and practice?

Judging whether a new care method really works (evidence-based practice).
Reading and appraising research articles critically (p values, CIs, tests used).
Planning and interpreting nursing research and the thesis or project.
Comparing outcomes between wards, groups or time periods.
Estimating community health indicators from surveys with a measure of precision.

Conclusion: Inference lets us move from the patients we measured to the patients we care about, while being honest about uncertainty.

Course references (as listed in the syllabus): Saunders, B. D., & Trapp, R. C. (1994). Basic and Clinical Biostatistics (2nd ed.). New Delhi: Prentice-Hall.

Source syllabus: Course Syllabus – LUMHS Post RN BSN Bio-Statistic (original post dated May 4, 2017).

Series (Biostatistics, Course No. 717): Part 1: Unit I – Introduction to Biostatistics. Part 2: Unit II – Scales of Measurement. Part 3: Unit III – Presenting Data. Part 4: Unit IV – Summarizing Data. Part 5: Unit V – Making Inference from Data. Part 6: Biostatistics for Nurses – facts, skills, uses, benefits and hurdles.

Independent educational resource: POST RN BSN Nursing Resources is not affiliated with, endorsed by or an official publication of LUMHS or PNMC. Model answers are study aids; verify against your course notes and textbooks.

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