Mathematics
Cass Business School Statistics ANOVA & P Value Questions

### Question Description

I need help with a Statistics question. All explanations and answers will be used to help me learn.

You do not need to solve Q1 and Q2.

I did Q1 and Q2 already

you have to do just Q3 and Q4 for me.

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Question 1 A number, π, of contestants are registered to take part in an archery contest. The distance between the centre of the target and the point that the π π‘β archerβs arrow hits is given by the random variable ππ , for π = 1, β¦ , π. The random variables π1 , β¦ , ππ are independent and identically distributed, each following an exponential distribution with a mean of 10cm. Each archer has one shot and the archer whose arrow hits closest to the centre of the target wins the contest. a) Determine the probability density function of the winnerβs distance from the centre of the target, that is, the density of the random variable π = πππ{π1 , β¦ , ππ }. (3 marks) b) Because of an outbreak of food poisoning in the hotel where the contestants are staying, it is possible that not all registered archers can participate in the contest. The number of archers taking part is given by the discrete random variable π, such that π(π = 10) = 0.8, π(π = 9) = 0.1, π(π = 8) = 0.1 and π is independent of π1 , π2 , β¦ Hence, the winnerβs distance from the centre of the target is given by the random variable π = πππ{π1 , β¦ , ππ }. Using the properties of conditional expectation, calculate πΈ(π). (4 marks) Total: 7 marks Question 2 Consider the random variables π1 and π2 , with means πΈ(π1 ) = πΈ(π2 ) = 0 and variances πππ(π1 ) = πππ(π2 ) = 1. The random variables follow the bivariate normal distribution, which means that their joint probability density function is given by 1 π₯12 β 2π₯1 π₯2 π + π₯22 π(π₯1 , π₯2 ) = ππ₯π {β }, π₯1 , π₯2 β β 2(1 β π2 ) 2πβ1 β π2 where π β (β1,1). You can take as given that each of π1 , π2 follows a standard normal distribution and that their correlation coefficient is π. a) Show that if π = 0, the random variables π1 , π2 are statistically independent. (1 mark) b) Show that the conditional density ππ1 |π2 takes the following form: 2 1 1 π₯1 β π₯2 π ππ1 |π2 (π₯1 |π₯2 ) = ππ₯π {β ( ) } 2 2 β1 β π2 β2πβ1 β π (1 mark) 1 c) For π = 0.99, state the value of πππ(π2 |π1 = π₯) for some π₯ and interpret it. d) Define the random variable π3 = (π1 all steps, and interpret your finding. )2 (2 mark) . Show that πΆππ£(π1 , π3 ) = 0, carefully justifying (4 marks) Total: 8 marks Question 3 Four university lecturers (A, B, D, and C) teach four modules each within a given academic year. The sample mean and variance of each lecturerβs module evaluation score, calculated across each lecturerβs modules, are given in the table below. Number of modules Lecturer A 4 Lecturer B 4 Lecturer C 4 Lecturer D 4 Average score Variance of scores 2.60 0.2196 3.13 0.3751 3.56 0.1851 3.92 0.2416 a) Perform a one-way Analysis of Variance for the above data, stating clearly the hypotheses tested and reporting your test result at the 5% significance level. (You may assume that all assumptions of the one-way ANOVA model are satisfied. You are given the following critical values of the F distribution, one of which will be needed to answer this question: πΉ3,12,0.025 = 4.474, πΉ12,3,0.05 = 8.745, πΉ3,12,0.05 = 3.490, πΉ4,12,0.05 = 3.259.) (4 marks) b) After being called in by his Head of Department to discuss his low feedback scores, Lecturer A claims that the reason his scores are comparatively low is that his class sizes were large. The following scatter-plot shows all lecturersβ scores plotted against the sizes of the four classes they each taught, together with the line of best fit, obtained via simple regression model of the form ππ = π½0 + π½1 π₯π + ππ , ππ βΌ π(0, π 2 ), where ππ are the evaluation scores for individual modules and π₯π are the corresponding class sizes. i. The estimate of the variance π 2 in the simple linear regression model is π  2 = 0.2744. Calculate the values of π 2 and of the correlation coefficient of the evaluation scores with the class size. (4 marks) 2 Figure 1 From the plot, estimate the value of the intercept, π1 . You are given that the standard error of π1 is π π΅1 = 0.0039. Calculate a 95% confidence interval and state your conclusion. (You may assume that the relevant critical value of the t distribution is approximately 2.) (3 marks) The Head of Department is not convinced that class size explains poor evaluation scores. She states that it may just be a coincidence that the worst performing lecturers teach larger classes. Explain what further analysis could be carried out to explore the issue further. (2 marks) Total: 13 marks ii. c) 3 Question 4 Let the random variable π represent the effort that a randomly chosen actuarial science student puts towards studying for a statistics module (on a scale from 0 to 5) and the random variable π represent that studentβs final exam mark. Assume that the conditional expectation of π given π = π₯ be given by the following formula: πΈ(π|π = π₯) = 20 + 10π₯ + 20 β tanh(π₯ β 2), where π 2π₯ β 1 tanh(π₯) = 2π₯ π +1 is the hyperbolic tangent function. The graph of the function π(π₯) = πΈ(π|π = π₯) is represented by the solid line in Figure 2 below. a) Let π be normally distributed, with mean equal to 2 and standard deviation equal to 0.5. Provide a simulation algorithm for calculating numerically the unconditional expectation πΈ(π), starting from π standard normal observations π§1 , β¦ , π§π . (5 marks) b) Calculate πΈ(π|π = 0) and interpret the result. (1 mark) c) An education researcher, who is not aware of the formula for πΈ(π|π = π₯) given above, tries to understand the relationship between studentsβ effort and their final exam mark. The researcher manages to collect data from (π, π) for 20 students. The data are shown in Figure 2 as points. The lecturer is fitting two regression models to the data, with prediction equations: Model 1: π¦Μ = β23.88 + 32.56 β π₯ Model 2: π¦Μ = 19.83 β π₯ Model 1 is a standard linear regression model. Model 2 is fitted by fixing the intercept to 2 zero, that is, 19.83 is the minimiser of the expression min β20 π=1(π¦π β π β π₯π ) . b Draw the regression lines corresponding to the two models on (a print-out of) Figure 2 and include this in your coursework submission. (3 marks) d) Comment on the rationale behind Model 2. Explain whether Model 1 or Model 2 is preferable. (3 marks) Total: 12 marks 4 Figure 2 5 6 ...
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QUESTION 3
a.

One-way Anova
H0:The mean score for each lecture modules are equal.
Ha: At least one of the mean score is different.
Mathematically it can be written as;
H0:Β΅1=Β΅2=Β΅3=Β΅4
Ha:At least one Β΅i is not equal tom 0.

Calculation of F-statistics

2.6+3.13+3.56+3.92
The overall mean is πΜΏ=
= 3.3025
4

Sum of square between groups,SSB= β4π=1 ππ (π₯Μπ β πΜΏ) = 4 Γ (2.6 β 3.3025)2 +
4 Γ (3.13 β 3.3025)2 + 4 Γ (3.56 β 3.3025)2 + 4 Γ (3.92 β 3.3025)2
=3.8835

Sum of square within group , SSE = β4π=1(ππ β 1) Γ π π2 = (4 β 1) Γ 0.2196 + (4 β 1) Γ
0.3751 + (4 β 1) Γ 0.1851 + (4 β 1) Γ 0.2416 = 3.0642

SST =SSB+SSE = 3.0642+3.8835= 6.9477

Df (between group) = k-1=4-1=3
Df(within group) =N-k =15-3=12
Total degree of fr...

psumanrec (4270)
UC Berkeley
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