Question

Could you help with these questions please

Activity Two: A Frequency Table Then Probability Table appliance store owner is interested in the relationship between price (retail or sales) the customer's decision to purchase an extended warranty or not. The owner defines the following events: A Item is purchased at retail price; C Extended warranty is purchased; B-ltem is purchased at sale price; D-Extended warranty is not purchased hen examining 2000 invoices he determined the following: 420 purchases made at the retail price also involved the purchase of extended warranty. A total of 160 purchases were made at the sale price but without extended warranty. In total, 700 purchases were made at the original retail price. (a) construct a bi-variate table, which represents the frequency of purchases for the relevant joint events and (b) construct a corresponding joint probability table, which includes the marginal probabilities. (Adapted from Selvanathan et al, 2000, p. 149) Activity Three: Independent Events and Contingency Analysis Referring to the previous activity, draw a table that represents the frequency for each joint outcome if the two events of interest were independent of each other. Comment on how close each value is between what actually happened and the expected outcome. Activity Five: Contingency Analysis and Chi-Square Test Using Excel, in your own time you may like to show that you can calculate a chi-square test statistic of 203.3 based on the tables you used in activity two and three. You won't be asked to do such extensive calculations in an exam, but could be asked to do so as part of an assignment using Excel. You should, however, know how to interpret this statistic in preparation for an examination style question. So, what conclusion would one make using this statistic? To do so, you need to consider the degrees of freedom and obtain a chi-square critical value to help you form a conclusion. Activity Four: Applying the Rules of Probability done to date. With these tables, any question becomes fairly straightforward to answer. following probabilities in words, and then find its actual value: Half the battle in answering probability questions is to generate the tables you have So, using the table(s) you generated in the two previous activities, express each of the (d) P(BID) (a) P(A) (e) P(CIA) (b) P(A) (f) P(A and C) (c) P(D | B) (8) P(A and D) (h) P(CIA)

Answer 1

2)

Frequency Table

3)

5)

Chi Square Test

H₀: The two variables (Price Type and Extended Warranty) are independent.

H_a: The two variables are associated.

Given,

Observed Values:

Retail Price (A Sale Price (B) Total 11401560 Extended Yes (C) Warranty PurchasedNo (D) 420 160 1300 280 440 Total 700

Expected Values:

Expected Values are calculated as:

E_ij = (T_i * T_j)/N

where,

T_i = Total in ith row

T_j = Total in jth column

N = table grand total

Alpha = 0.05

df = (r-1)*(c-1) = (2-1)*(2-1) = 1*1 = 1

Chi Square _Critical = 3.84

Decision Rule:

If Chi Square> Chi Square _Critical reject the null hypothesis

Test Statistic:

Chi Square = ∑(O_ij – E_ij)²/E_ij = (420-546)²/546 + ……………….. + (160 – 286)²/286 = 203.34

Result:

Since, Chi Square> Chi Square _Critical ­­we reject the null hypothesis.

Conclusion:

Price Type and Extended Warranty are associated.

4)

Using the frequency from part 2

Probability = Favourable Outcome/Total Outcome

a)

P(A) = 420/2000 = 0.21 or 21%

b)

P(A^C) = 1 – P(A) = 1-0.21 = 0.79 or 79%

c)

P(D|B) = 160/1300 = 0.1231 or 12.31%

d)

P(B|D) = 160/440 = 0.3636 or 36.36%

e)

P(C|A) = 420/700 = 0.60 or 60%

f)

P (A and C) = 420/2000 = 0.21 or 21%

g)

P (A and D) = 280/2000 = 0.14 or 14%

h)

P(C|A^c) = 0