Basic Statistics For The Health Sciences Homework Chapter 55

Basic Statistics For The Health Scienceshome Work Chapter 554 A Coupl

Basic Statistics for the Health Sciences Home Work chapter .4 A couple is planning to have three children. Find the following probabilities by listing all possibilities and using equation 5:

  • a) 2 boys and one girl
  • b) At least one boy
  • c) No girls
  • d) At most 2 girls
  • e) Two boys followed by a girl

5.10 If an individual were chosen at random from table 2.1, what is the probability that that person would be:

  • a) vegetarian
  • b) a female
  • c) a male vegetarian

Chapter .6 If the heights of male youngsters are normally distributed with a mean of 60 inches and a standard deviation of 10, what percentage of the boys' heights (in inches) would we expect to be:

  • a) between 45 and 75
  • b) between 30 and 90
  • c) less than 50
  • d) 45 or more
  • e) 75 or more
  • f) between 50 and

a) Suppose that 25-year-old males have a remaining mean life expectancy of 55 years with a standard deviation of 6. What portion of 25-year-old males will live past 65?

b) What assumption do you have to make in order to obtain a valid answer?

Chapter .6 Suppose heights of 20-year-old men are approximately normally distributed with a mean of 71 inches and a population standard deviation of 5 inches. A random sample of 15 20-year-old men is selected and measured. Find the probability that the sample mean:

  • a) is at least 77 inches
  • b) lies between 65 and 75 inches
  • c) is not more than .16 if the forced vital capacity of 11-year-old white males is normally distributed with a mean of 2400cc and SD of 400, find the probability that a sample of size n=64 will provide a mean:
  • a) greater than 2500
  • b) between 2300 and 2500
  • c) less than 2350

Paper For Above instruction

The probability calculations in health sciences often involve understanding basic statistical concepts such as probability distributions, hypothesis testing, and confidence intervals. This paper explores several probability problems relevant to health sciences, emphasizing understanding from simple events to more complex statistical inference.

Probabilities in Family Planning

In the context of family planning, calculating probabilities involves enumerating all possible outcomes. For a couple planning to have three children, each child can be a boy (B) or a girl (G), giving 2 options per child. The sample space consists of 2^3 = 8 outcomes: {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}.

a) The probability of having exactly two boys and one girl is obtained by counting outcomes like BBG, BGB, GBB, totaling 3 favorable outcomes. Since all outcomes are equally likely, the probability is 3/8.

b) Probability of at least one boy includes all outcomes except GGG, totaling 7/8.

c) No girls occurs only in the GGG outcome, so probability is 1/8.

d) At most 2 girls encompasses all outcomes except GGG, hence probability is 7/8.

e) The probability of two boys followed by a girl (sequence: BBG) is 1/8, assuming each child's gender is independent with equal probability.

Probability from Population Data

When selecting an individual randomly from a dataset, probabilities depend on the composition of the population.

Suppose table 2.1 contains data on dietary habits and gender. If, for instance, out of 200 individuals, 50 are vegetarians, then the probability of selecting a vegetarian person is 50/200 = 0.25.

  • a) The probability that a randomly chosen person is vegetarian is determined by dividing the number of vegetarians by the total number of individuals.
  • b) The probability that the person is female is based on the female count divided by the total sample size.
  • c) The probability that the person is both male and vegetarian is proportional to the number fitting both categories divided by the total sample size.

Normal Distribution and Height Percentages

Height of male youngsters is normally distributed with a mean (μ) of 60 inches and standard deviation (σ) of 10 inches. To determine percentages, standard normal distribution (Z) scores are used.

For example, the percentage within 45 to 75 inches is calculated by converting these to Z-scores:

  • Z for 45 = (45 - 60)/10 = -1.5
  • Z for 75 = (75 - 60)/10 = 1.5

Using standard normal tables, the area between Z = -1.5 and Z = 1.5 is approximately 86.64%, indicating that about 86.64% of boys' heights fall within this range.

Similarly, percentages for other ranges are calculated by converting bounds to Z-scores and using standard normal tables. For example, less than 50 inches (Z = (50-60)/10 = -1), which corresponds to about 15.87% of the population.

Life Expectancy and Normal Approximation

For 25-year-old males, with mean life expectancy of 55 years and SD of 6 years, the question pertains to the proportion living past 65. First, find Z for 65:

Z = (65 - 55)/6 ≈ 1.67

Using the standard normal table, the probability of living past 65 (Z > 1.67) is approximately 0.0475 or 4.75%. This implies a small proportion of 25-year-old males will live beyond 65, assuming normality.

The assumption necessary here is that the life expectancy distribution is approximately normal, which allows use of the standard normal distribution to estimate probabilities.

Sampling Distributions and Probabilities

When sampling from a population, the distribution of the sample mean depends on the population distribution and the sample size. For heights of 20-year-old men with mean 71 inches and SD 5 inches, the sampling distribution of the mean has a standard error of:

  • SE = σ/√n = 5/√15 ≈ 1.29 inches.

a) To find the probability that the sample mean is at least 77 inches, convert to Z:

Z = (77 - 71)/1.29 ≈ 4.65

From Z-tables, the probability is practically zero, indicating this event is highly unlikely.

b) The probability that the sample mean lies between 65 and 75 inches involves calculating Z-scores and the corresponding areas, which can be obtained from standard normal tables.

c) For the probability that the sample mean is not more than 0.16, similar Z-score and table calculations are used.

Probability of Sample Means in Normal Distributions

Considering the forced vital capacity (FVC) with mean 2400cc and SD 400cc, the probability that a sample of size 64 has a mean greater than 2500cc requires calculating the standard error:

SE = 400/√64 = 50cc

Z = (2500 - 2400)/50 = 2

The probability of a sample mean exceeding 2500cc is the area to the right of Z=2, approximately 0.0228.

Similarly, probabilities for other ranges are calculated by converting the bounds to Z-scores.

Hypotheses Testing in Environmental and Health Studies

Formulating hypotheses involves setting null and alternative statements. For example:

  • a) Null (H0): The mean community level of suspended particulates is 30 units.
  • Alternative (H1): It exceeds 30 units.

Statistical tests determine if observed data are consistent with H0, typically using significance levels. A similar process applies to testing the mean age of disease onset, IQ scores, and other health parameters, where null hypotheses represent no effect or no difference, and alternative hypotheses imply the presence of an effect or difference.

Population Comparisons and Proportions

In comparing proportions between two populations, such as Ecuadorian villagers versus U.S. populations, statistical tests evaluate whether observed differences are statistically significant. For instance, with 29 out of 99 Ecuadorian individuals aged 65+ and 20% in the U.S., a hypothesis test can determine if Ecuador's proportion exceeds that of the U.S., considering the significance level (α = 0.01).

Estimating Population Proportions with Confidence Intervals

When estimating the proportion of men infected with AIDS in a small community, the point estimate is directly from the data: 13/100 = 0.13. To calculate the 95% confidence interval, the standard error (SE) is computed:

SE = √(p̂(1 - p̂)/n) = √(0.13*0.87/100) ≈ 0.033

The critical z-value for 95% CI is approximately 1.96, so the interval is:

0.13 ± 1.96*0.033 ≈ (0.065, 0.195)

This interval suggests that the true proportion of infected men is between 6.5% and 19.5% with 95% confidence.

Conclusion

Understanding and applying basic statistical concepts such as probability calculations, normal distribution properties, hypothesis testing, and confidence intervals are essential in health sciences. These tools enable researchers to make informed decisions, evaluate risks, and establish evidence-based practices. Accurate interpretation of statistical data contributes significantly to public health policies, epidemiological surveillance, and clinical research, ultimately improving health outcomes.

References

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