A comparative study compares two or more groups to establish whether there are statistically significant differences between the groups with respect to some parameter(s).

Examples include control group and test group (controlled store tests), users and non-users of a brand and before-and-after analysis.

The study tests the null hypothesis (H_{0}) that the groups exhibit no differences in responses.
The alternative or research hypothesis (H_{A}) represents a difference between groups, i.e., rejection of the status quo.

In the context of sampling, given the non-null difference that the analyst wants to detect, we need to know the sample size required to come to a conclusion at specified levels of accuracy.

In addition to α and β (type I and type II errors), the sample size estimate depends on the standard deviation of the population and the smallest effect or non-null difference that the analyst wants to detect.

The formula for sample size for two groups of equal sample sizes:

$$ n = \frac {2 (z_α + z_β)^2 }{(δ/σ)^2} $$δ = |μ_{0} − μ_{1}| is the detectable difference in the mean.

σ: population variance.

z_{α}: standardized value associated with α, the level of significance.

z_{β}: standardized value associated with β, the type II error.

For α = .05, z_{α} = 1.96; for β = .20, z_{β} =0.84.

Δ is the standardized difference between means, or the effect size (ES). It is measured in units of the standard deviation and represents the magnitude of difference in means.

The multiplier, M, varies with power as depicted in Exhibit 34.7.

Power (1 – β) | One sample | Two sample | |
---|---|---|---|

0.5 | 4 | 8 | |

0.8 | 8 | 16 | |

0.9 | 11 | 21 | |

0.95 | 13 | 26 | |

0.975 | 16 | 31 |

The detectable difference, |μ_{0} − μ_{1}|= σ√(M/n)

For the conventional setting of power = 0.8, M = 16:

$$ n=\frac{16}{∆^2}, |μ_0- μ_1|=4σ/\sqrt{n} $$This is the sample size for *each* of the two groups. If there is only one sample,
the multiplier is 8, and n = 8/Δ^{2}.

Consider, for example, a study on the weight of fresh recruits in the army. If we are interested in detecting a change in weight over a previous batch of recruits, and the magnitude of the difference of practical relevance is 400gm or more, and the standard deviation is 3.2 kg, in that case:

$$ ∆=\frac{400}{3200}=0.125; \;n=\frac{8}{∆^2} =512 $$We need a sample of 512 recruits to detect a difference of 400gm in weight.

For paired observations such as before and after, the sample size formula is:

$$ n = \frac{(z_α + z_β)^2}{(δ/σ)^2} $$Note the multiplier, M, has been halved, and, importantly, σ is the standard deviation of the differences within pairs, which is rarely known in advance.

The sample size required to compare two proportions, p0 and p1:

$$ n = \frac{M p(1 - p)}{(p_0 - p_1)^2} \;\;\;where \; M \;is \;the \;multiplier \;and \;p = (p_0+ p_1)/2 $$Taking the conservative estimate p = 0.5 results in upper limit on the required sample size, and if power is set at the conventional level of 0.8, the multiplier M is 16, then:

$$n = \frac {M}{4×(p_0 - p_1)^2} $$ $$n = \frac {4}{(p_0 - p_1)^2} $$For small proportions, p<.05, use: n = 4/(√p_{0} - √p_{1})^{2}

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