Paired group test with unknown standard deviation is essentially the same as a single sample t-test. The paired values are reduced to a single series by computing the difference between the two sets.
Example: In advertising copy tests, purchase intent is gauged through pre-post exposure measurement of respondents’ disposition to buy or try a product. The metric, x, is usually the top-2 box rating (“very likely to buy”, “likely to buy”). The paired values (x_{before}, x_{after}) are reduced to a single set of values (y) by computing the difference: y = x_{after} – x_{before}. H_{A}: μ > 0, i.e., exposure to the advert is expected to improve disposition to try the product.
Example: Sequential monadic tests are frequently used for product testing. The respondents try one product and rate it, move to another product and rate it, and then compare the two. A paired t-test may be used to determine whether an improved formulation is rated higher on an attribute. H_{A}: μ > 0, i.e., new product expected to be rated higher.
The remaining steps are the same as that for a one-tailed, single sample t-test.
Comparison test for two groups of unknown standard deviation also requires use of the t-test, since the population’s standard deviation is not known.
Example: A study was conducted to examine the consumption of coffee by office workers. The statistics for the men and women sampled in this study are given below.
Men: Sample size n_{M}=440, mean x̄_{M} =46.5 cups per month, standard deviation s_{M}=36.3.
Women: Sample size n_{W}=360, mean x̄_{W}=35.1 cups per month, standard deviation s_{W}=20.6.
There is a difference of 11.4 cups per month in coffee consumption between men and women. Is this difference resulting from sampling error, or do men consume significantly more coffee than women?
H_{0}: μ_{M} – μ_{W} < 0
H_{A}: μ_{M} – μ_{W} > 0
α = 0.05
Standard deviation:
$$ σ_{\bar x_M-\bar x_M} = \sqrt {\frac{s_M^2}{n_M} + \frac{s_W^2}{n_W}} = \sqrt {\frac{36.3×36.3}{440} + \frac{20.6×20.6}{360}} = 2.04 $$ $$ t=\frac{\bar x_M-\bar x_W}{σ_{\bar x_M - \bar x_W}} =\frac {46.5-35.1}{2.04}=5.58 $$ $$ degrees\;of\; freedom\; (df) = \frac {\left(\frac {s_M^2}{n_M} + \frac {s_W^2}{n_W}\right)^2} {\frac{(s_M^2/n_M)^2}{n_M-1}+\frac {(s_W^2/n_W)^2}{n_W-1}} = 716.8 $$p-value = 0.00001 obtained from t distribution.
The probability of obtaining a t-value of 5.58 or higher with 717 degrees of freedom, when sampling 440 men and 360 women is very low (0.00001 << α = 0.05). More specifically, if women were consuming as much coffee as men, the chance that the sample differences would average 11.4 or more cups per month is only 0.00001. The null hypothesis is rejected. The data strongly suggests that women consumers consume less coffee than men.
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