Question

(c) A sample of 80 people is selected at random from a demographic group to complete a questionnaire on lifestyle choices. As part of this survey, each respondent is asked whether he or she regularly take part in organised sport. It is known that 29% of all the people in the demographic group regularly take part in organised sport. Use a suitable approximation to calculate the probability that at least 20 of the people in the sample regularly take part in organised sport. (6 marks) (d) Assume that you are using a test to check whether the residuals of a regression model can be considered to come from a normal distribution. You assume that the mean of the population of residuals is zero because of the way in which the residuals are calculated. Using the sample data, you have estimated the standard deviation of the population of residuals as the sample standard deviation s, which you calculate as 10.9; you have classified the residuals into classes as follows less than-11 13 13 20 25 14 21 1 to -5 5 to 0 0 to 5 5 to 11 more than 11 Assuming that that the residuals are normally distributed: (i) Calculate the z values of the class limits. (ii) Calculate the probability that a standard normal variable will fall in each class. (iii) Find the expected frequencies in each class. (iii)Calculate a statistic to test the normality of the population from which the (iv)State the degrees of freedom of the statistic. (vi) State whether the test statistic gives evidence at the 5% level that would make residuals come. you reject the assumption of normality. (13 marks) (Total 25 marks)

Answer 1

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n ………………................................................................…..(1)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST…………………………...........................................................................................................…………….(1a)

Mean (average) of X = E(X) = µ = np……………….............................................................………………………………..(2)

Variance of X = V(X) = σ² = np(1 – p)……………………….......................................................…………………………..(3)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)} …….........................................…………………………....………...(4)

If X ~ B(n, p), np ≥ 10 and np(1 - p) ≥ 10, then Binomial probability can be approximated by Standard Normal probabilities by Z = (X – np)/√{np(1 - p)} ~ N(0, 1) …...............................................................…………...................................………..(5)

Now, to work out the solution,

Part (c)

Let X = number of respondents in the sample of 80 people who regularly take part in organised sport. Then, X ~ B(80, p), where p = proportion of all people in the demographic region who regularly take part in organised sport and this is given to be 0.29 (i.e., 29%). Thus, X ~ B(80, 0.29) ......................................................................................................................... (6)

Now, probability at least 20 people in the sample regularly take part in organised sport

= P(X ≥ 20)

= P[Z ≥{(20 – 23.2)/4.015}] [vide (5) and (6)]

= P(Z ≥ - 0.7970]

= 0.7813 [Using Excel Function: Statistical NORMSDIST ] Answer 1

Part (d)

All details are presented in the table below, in which the following codes are used just for ease in presentation.

CLi: Class Number

LB: Lower Boundary

UB: Upper Boundary

O: Observed Frequency

ZU: Z-value of UB = {(UB – 0)/10.9}

Φ: Cumulative Probability of N(0, 1) [Using Excel Function: Statistical NORMSDIST ]

pCL: Probability of the class

E: Expected Frequency = 106 x pCL

χ² : {(O – E)²/E}

Cli	Lbi	Ubi	Oi	Zui	Φ(ZU)i	pCLi	Ei	χ2
1		-11	13	-1.00917	0.1564	0.15645	16.58	0.7742
2	-11	-5	13	-0.45872	0.3232	0.16677	17.68	1.2379
3	-5	0	20	0	0.5	0.17678	18.74	0.0849
4	0	5	25	0.458716	0.6768	0.17678	18.74	2.0921
5	5	11	14	1.009174	0.8436	0.16677	17.68	0.7652
6	11		21	6	1	0.15645	16.58	1.1764
		Total	106			1	106	6.1307
Sub-parts				(i) Ans 2		(ii) Ans 3	(iii) Ans 4	(iv) Ans 5

Sub-part (v) : Degrees of freedom = Number of Classes – Number of Parameters estimated (only SD)

= 6 – 1

= 5 Answer 6

Sub-part (vi)

Decision

At 5% significance level, χ²_crit= 11.07 [Using Excel Function: Statistical CHIINV ]

From the above table, χ²_obs= 6.13.

Since χ²_obs< χ²_crit, null hypothesis of Normality cannot be rejected. Answer 7

DONE