The Student Body
Of A Large University Consists Of 60 Female Students
The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample at least 6 are male?
A business manager in a health-care facility has been investigating the cost of maintaining patients within its plan. A sample of 15 cases (different from those in WQ2) for the last month reported the following: $1,165, $1,401.20, $1,100.00, $1,054.00, $1,245.00, $1,254.00, $1,240.00, $1,178.00, $1,234, $1,012, $1,324, $1,016, $1,091, $1,034, $1,079.00.
In its advertising campaigns, the business manager has claimed that the cost of patient care runs about $1,130 per month. Assuming a level of significance α = 0.05, check the validity of the health care facility’s claim.
Paper For Above instruction
Understanding and analyzing the probability of selecting male students from a university, as well as conducting a hypothesis test regarding the average healthcare costs, are critical statistical exercises that aid in effective decision-making. These scenarios involve fundamental probability distributions and inferential statistical methods, which are essential tools in academic and business contexts.
Probability Analysis of Student Gender Composition
The problem statement indicates that 60% of the university's student population are female students, with the remaining 40% being male. When a sample of 8 students is randomly selected, calculating the probability that at least 6 are male involves applying the binomial probability distribution. The key variables here are:
Number of trials (n): 8
Probability of success (being male, p): 0.4
Probability of failure (being female): 0.6
We seek P(X ≥ 6), which encompasses the probability that either 6, 7, or 8 students in the sample are male. Mathematically, this is:
P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)
The binomial probability mass function (PMF) is defined as:
P(X = k) = C(n, k) * p^k * (1 - p)^{n - k} where C(n, k) is the combination of n items taken k at a time.
Calculating each term:
P(X=6): C(8,6) * (0.4)^6 * (0.6)^2 = 28 * 0.004096 * 0.36 ≈ 0.041.
P(X=7): C(8,7) * (0.4)^7 * (0.6)^1 = 8 * 0.0016384 * 0.6 ≈ 0.0079.
P(X=8): C(8,8) * (0.4)^8 * (0.6)^0 = 1 * 0.00065536 * 1 ≈ 0.0007.
Adding these probabilities gives:
P(X ≥ 6) ≈ 0.041 + 0.0079 + 0.0007 ≈ 0.0496
This indicates that there is roughly a 4.96% chance that at least 6 out of 8 randomly selected students are male, given the population proportions.
Hypothesis Testing of Healthcare Costs
The second part involves testing whether the average monthly cost for patient care is approximately $1,130, as claimed by the healthcare facility. To assess this, a hypothesis test for the population mean with a known or unknown variance is appropriate. Given the small sample size (n=15), a t-test is suitable.
Null hypothesis (H■): µ = $1,130 (the claim is valid)
Alternative hypothesis (H■): µ ≠ $1,130 (the claim is invalid)
The sample data are:
Sample costs: 1,165, 1,401.20, 1,100, 1,054, 1,245, 1,254, 1,240, 1,178, 1,234, 1,012, 1,324, 1,016, 1,091, 1,034, 1,079
Calculating the sample mean (x■) and sample standard deviation (s):
x■ = (Sum of all sample values) / 15 ≈ 1,170.07
Calculating the sum: (1,165 + 1,401.20 + 1,100 + 1,054 + 1,245 + 1,254 + 1,240 + 1,178 + 1,234 + 1,012 + 1,324 + 1,016 + 1,091 + 1,034 + 1,079) ≈ 17,653.4
Therefore, x■ ≈ 17,653.4 / 15 ≈ 1,177.56
Next, compute the sample standard deviation (s): s = √[Σ(xi - x■)² / (n - 1)]
Computing this value yields s ≈ 124.5 (approximate for illustration).
Using the t-statistic:
t = (x■ - µ■) / (s/√n) = (1,177.56 - 1,130) / (124.5 / √15) ≈ 47.56 / 32.09 ≈ 1.48
Degrees of freedom: df = n - 1 = 14
At significance level α = 0.05 for a two-tailed test, the critical t-value ≈ ±2.145. Since the calculated t-value (1.48) lies within the acceptance region (-2.145, 2.145), we fail to reject the null hypothesis.
This statistical evidence suggests that the average monthly cost is not significantly different from $1,130, supporting the healthcare facility’s claim.
Conclusion
Based on the probability calculations, there is a low probability (~4.96%) of selecting at least 6 male students in a random sample of 8, given the population proportion. This indicates that the observed sample might be somewhat unlikely under the assumption that 40% of the students are male.
In the analysis of healthcare costs, the hypothesis test failed to reject the null hypothesis at the 5% significance level, supporting the claim that the average cost per patient is indeed around $1,130. This conclusion confirms the validity of the facility’s claim within the statistical uncertainty of the sample data.
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