Question

A newly opened local restaurant has just finished a media blitz (Instagram, email, mobile, etc.) for new customers. Whenever prospective customers visited the restaurant the first time, they were asked to indicate what ads they saw in the past week. Some of these prospects came back and became the frequent customers, while some did not; thus, we have two populations: those who were the regular customers of this restaurant and those who were not. At the end of the 7-day campaign, a staff member performed the following tabulations.

Joined the Health Club Did Not Join the Health Club

Total visitors 80 35

Recalled Instagram ads 55 20

Recalled email ads 48 30

Recalled mobile ads 72 11

1. Use your knowledge of the formula and the test of the significance of the difference between two percentages to ascertain if there are any significant differences in this data. What are the implications of your findings with respect to the effectiveness of various advertising media used during the membership recruitment ad blitz?

Answer 1

Solution

Solution is based the Z-test for equality of population proportions.

For each of the three media, test will be performed on the recalling proportion of those who Joined the Health Club versus those Did Not Join the Health Club.

Let p₁ be population recalling proportion for those who Joined the Health Club and p₂ be population recalling proportion for those who Did Not Join the Health Club.

Let p_1cap and p_2cap be the corresponding sample proportions.

Back-up Theory

Z-test for equality of population proportions.

Hypotheses:

Null H₀ : p₁ = p₂   Vs H_A : p₁≠ p₂

Test Statistic:

Z = (p₁cap – p₂cap)/√[pcap(1 - pcap){(1/n₁) + (1/n₂)} where p₁cap and p₂cap are sample proportions, n₁, n₂ are sample sizes and pcap = {(n₁ x p₁cap) + (n₂ x p₂cap)}/(n₁ + n₂).

Distribution, Significance Level α, Critical Value :

Under H₀, distribution of Z can be approximated by Standard Normal Distribution

So, given a level of significance of α%, Critical Value = upper (α/2)% of N(0, 1), and

Using Excel Functions of N(0, 1), Statistical NORMSINV Critical Value is found.

Decision Criterion (Rejection Region):

Reject H₀, if | Zcal | > Zcrit .

Now to work out the solution,

For all tests, Significance Level α is taken to be 5% [i.e., 0.05] and hence Critical Value = upper 2.5% of N(0, 1) = 1.96

{(1/n₁) + (1/n₂)} = (1/80) + (1/35) = 0.0411

Case 1: Media: Instagram

Here p_1cap = 55/80 = 0.6875

p_2cap = 20/35 = 0.5714 and hence

pcap = 75/115 = 0.6522

So, Z = 0.1161/√(0.6522 x 0.3478 x 0.0411)

= 1.2024

Since Z < Zcrit, H₀ is accepted and hence we conclude that there is not enough evidence to conclude that the recall proportion is different between members and non-members of the club.

Case 2: Media: email

Here p_1cap = 48/80 = 0.6

p_2cap = 30/35 = 0.8571 and hence

pcap = 78/115 = 0.6783

So, Z = 0.2571/√(0.6783 x 0.3217 x 0.0411)

= 2.7148

Since Z > Zcrit, H₀ is rejected and hence we conclude that there is enough evidence to conclude that the recall proportion is different between members and non-members of the club.

Case 3: Media: Mobile

Here p_1cap = 72/80 = 0.9

p_2cap = 11/35 = 0.3143 and hence

pcap = 83/115 = 0.7217

So, Z = 0.5857/√(0.7217 x 0.2783 x 0.0411)

= 6.4464

Since Z > Zcrit, H₀ is rejected and hence we conclude that there is enough evidence to conclude that the recall proportion is different between members and non-members of the club.

Thus, overall conclusion would be:

1. recall proportion does depend on the media

2. recall proportion differs from members to non-members in two out of 3 media.

Answer