Question

marks) Ben runs a sports program for high school students. Each student that ooses one of the following three sports: volleyball, basketball and netball. The heights of the first 15 students, and the sports they chose, were Volleyball 183, 176, 170, 179 Basketball 165, 169, 171, 154, 165, 159 Netball 167, 160, 177, 173, 170 Ben runs an analysis of variance with these data. A partially complete ANOVA table from his analysis is given below. Source df SS MS F Treatment (sport) Error 38 Total 872.4 (a) Is there evidence of a relationship between the students' heights and their choice of sport? (b) what is the sampling distribution of σ2? e) Let μί be the population rnean of students who choose volleyball, what is the sampling distribution of 1? (d) Let X" be the height of the next student who chooses volleyball. Show that, x*-n-N(0,5の 2 (e) Calculate a 95% prediction interval for X*.

Answer 1

Solution

Preparatory Work

ANOVA Table Completed

Degrees of Freedom: Since there are 15 observations in total, df for Total is (15 - 1) = 14, There are 3 sports (treatments) and so the df = (3 - 1) = 2. Error df = 14 – 2 = 12.

MS = SS/df. So, Error SS = 38 x 12 = 456 and hence, Treatment SS = TSS – ESS = 872.4 – 456 = 416.4.

F = MST/MSE

Source	df	SS	MS	F	p-value
Treatment	2	416.4	208.2	5.4789	0.0204*
Error	12	456	38
Total	14	872.4	-

*p-value = P(F_2,12 > 5.4789)

Part (a)

Taking α = 5% (i.e., significance level), p-value < α. Hence, the null hypothesis of no difference in means is rejected.

Conclusion: There is sufficient evidence to suggest that the mean height of students is related to the choice of sport. Answer

Part (b)

Sampling distribution of σ²_cap is χ²₁₂, i.e., Chi-square with 12 degrees of freedom. Answer

Part (c)

A basic assumption of ANOVA is that the population is Normally distributed. Taking the population standard deviation to be σ, the sampling distribution of sample mean based n observation is N(µ, σ²/n).

Since Volleyball has 4 observations, n = 4 and hence sampling distribution of sample mean, µ_1cap is

N(µ, σ²/4). Answer

Part (d)

Since X* is also coming from the same population of students, X* ~ N(µ, σ²). From Part (c), µ_1cap ~ N(µ, σ²/n). So, (X* - µ_1cap) ~ N[(µ - µ), {σ² + (σ²/4)}] i.e., N(0, 5σ²/4) Answer

Part (e)

95% prediction interval for X* is: µ_1cap ± 1.96σ_cap, where

1.96 = upper 2.5% point of N(0, 1),

µ_1cap = mean height of ‘volleyball’ students = 177

σ_cap = √MSE = √38 = 6.1644

So, 95% prediction interval for X* is: [164.92, 189.08] Answer

DONE