A Random Variable X Has A Binomial Distribution B64 05 Use T
A Random Variable X Has A Binomial Distribution B64 05 Use The
A. Random variable X has a binomial distribution, B(64, 0.5). Use the normal approximation to compute P{26 ≤ X
B. Random variable X has a normal distribution, N(50, 100). Compute P{X
C. Test the claim that the population of Freshman college students has a mean grade point average greater than 2.00. Sample size = 24, sample mean = 2.35, sample standard deviation = 0.70, with a Type I error of 0.01. Include the test statistic, critical value(s), and conclusion.
D. Find a linear (regression) equation with the following data: x and y values.
E. Four groups, A, B, C, and D, were randomly selected from a normally distributed population. Test whether all four group means are equal by constructing an ANOVA table with sources of variation and appropriate degrees of freedom, sums of squares, mean squares, and F-value.
Paper For Above instruction
The statistical analysis of various probability distributions and hypothesis testing forms a fundamental component of statistical inference. This paper addresses five specific statistical problems involving binomial distributions, normal distributions, hypothesis testing, regression analysis, and analysis of variance (ANOVA). Each problem is approached with appropriate statistical methods, calculations, and interpretation of results, providing a comprehensive understanding of applied statistics techniques.
Problem A: Normal Approximation to Binomial Distribution
The first problem involves the binomial distribution B(64, 0.5). When the sample size is large, the binomial distribution can be approximated by a normal distribution under certain conditions, specifically when both np and n(1-p) are greater than 5. In this case, n=64, p=0.5, thus np=32 and n(1-p)=32, satisfying the condition for normal approximation.
The mean (μ) and standard deviation (σ) of the binomial distribution are calculated as:
μ = n p = 64 0.5 = 32σ = sqrt(n p (1 - p)) = sqrt(64 0.5 0.5) = sqrt(16) = 4
Applying the normal approximation, P{26 ≤ X
Standardizing these bounds:
Z1 = (25.5 - 32) / 4 = -1.625Z2 = (33.5 - 32) / 4 = 0.375
Using standard normal tables or computational tools, the probabilities are:
P(ZP(Z
Thus, the probability is approximately:
P{26 ≤ XAnswer: approximately 0.594.
Problem B: Probability from a Normal Distribution
Given X is normally distributed, N(50, 100), which implies a mean μ=50 and a variance σ²=100, hence standard deviation σ=10, we need to compute P{X
Expressing as the sum of probabilities:
P{XCalculating each with standardization:
Z1 = (45 - 50) / 10 = -0.5Z2 = (65 - 50) / 10 = 1.5
Using standard normal distribution tables:
P(ZP(Z > 1.5) = 1 - P(Z
Adding both probabilities:
0.3085 + 0.0668 ≈ 0.3753Answer: approximately 0.375.
Problem C: Hypothesis Test for the Mean GPA of Freshman Students
The claim that the mean GPA exceeds 2.00 is tested using sample data: n=24, \(\bar{x}=2.35\), s=0.70, with a significance level α=0.01.
The null hypothesis \( H_0: \mu \leq 2.00 \) and the alternative hypothesis \( H_1: \mu > 2.00 \).
Since the sample size is less than 30 and the population standard deviation is unknown, a t-test is appropriate:
t = (\(\bar{x}\) - μ₀) / (s / sqrt(n))= (2.35 - 2.00) / (0.70 / sqrt(24))
= 0.35 / (0.70 / 4.898)
= 0.35 / 0.1429 ≈ 2.45
Degrees of freedom df = 23. Using t-distribution tables or software, the critical t-value for α=0.01 in a one-tailed test:
t_critical ≈ 2.508Since the calculated t=2.45 is less than t_critical=2.508, we fail to reject the null hypothesis at the 1% significance level.
Conclusion: There is insufficient evidence at the 1% level to conclude that the average GPA exceeds 2.00.
Problem D: Linear Regression Equation
Given paired data for variables x and y, the regression line \( y = a + b x \) can be determined via least squares estimation. The formulas are:
b = Σ(xy) - n x̄ ȳ / Σ(x²) - n * x̄²a = ȳ - b * x̄
Without specific data points, the general approach involves calculating means, sums of products, and sums of squares, then solving for coefficients. For example, with data points (x₁, y₁), (x₂, y₂), ..., (x_n, y_n), these values are computed and plugged into the formulas to obtain the linear equation, which best fits the data in least squares sense.
Suppose the data points yield the following estimates: b = 0.75 and a = 1.2. The regression line then is:
y = 1.2 + 0.75 * xProblem E: ANOVA Table for Comparing Four Group Means
The goal is to test whether four groups A, B, C, and D have the same mean. The null hypothesis:
H₀: μ_A = μ_B = μ_C = μ_DThe ANOVA procedure involves calculating the sums of squares between groups (SSB), within groups (SSE), and total (SST), then dividing by the respective degrees of freedom to obtain mean squares, and calculating the F-statistic.
Suppose the data yields the following sums of squares and degrees of freedom:
Source df SS MS F
Total N-1
Groups 3 SSB
Error N-4 SSE
Where N is total number of observations across all groups. The F-value is computed as MS between / MS within. Decision rules depend on the critical F-value at chosen significance level, say 0.05. If F calculated > F critical, reject H₀. Otherwise, fail to reject H₀, indicating no significant difference among group means.
In practice, software like SPSS, R, or SAS provides these calculations, but essential formulas and principles remain the same. The interpretation guides whether differences in group means are statistically significant or not.
Conclusion
This comprehensive analysis demonstrates critical statistical methodologies for different distributions and tests. Using normal approximations, hypothesis testing, regression analysis, and ANOVA, researchers can infer meaningful insights from data and make data-driven decisions. Each problem illustrates fundamental statistical principles, emphasizing the importance of correct assumptions, calculations, and interpretations in applied statistics.
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