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BUAD 300Prof. RobinsonSign TestThe sign test is a non-parametric procedure often used to discern preferences among two possibilities. For example, it can be used to discern if consumers prefer brand X or brand Y. It is based on the binomial model that students review in their 200-level statistics course – the coin flip problem. It reduces to a standard normal z test when the number of observations is large. This nonparametric test is explained well on pages 872-878 of the 12th edition of your text. In the sign test, we survey a sample, and record one of two symbols such as a “+” or “-.” Allow the following:n = sample size – number who express “no preference.”p = the proportion of n who express a preference.u = the actual number who express a “+” preference (or some other preference).For this procedure, if 200 people are surveyed about their preference for brand X versus brand Y, but 30 indicate they have no preference, then n = 200 – 30 = 170. Consider a test of the following hypothesis:H0: The population has no preference, i.e. p = ?.Following the text (see page 876), we haveE(u) = n/2σu = n/4 If n > 30, then u is approximately normal, so that we have z = u-E(u)σu Review question: In a sample of 100 consumers concerning their preference for brand X versus brand Y, 20 expressed no preference, and 55 expressed a preference for X. Can we conclude that the overall population has no preference for X or Y? Let α = 5%.H0: There is no preference, i.e. if p is the proportion that prefers either X or Y, then p = 1/2.n = 100 – 20 = 80u = 55E(u) = n/2 = 80/2 = 40σu = n/4 = 20 = 4.47z calc = u-E(u)σu = 55-404.47 = 3.35Note that this is a two-tailed test since a population preference could result in either more than 40, or less than 40, expressing a preference for brand X or Y.Zcritical at α/2 = 2.5% = 1.96Decision rule: If |z calc.| > zcritical then reject H0.Decision: Reject H0!Assigned problems: Do problems 6 and 7 on page 879 of your text.Derivation of the Sign TestIf we have n binomial trials such as a coin flip, where the probability of either outcome symbolized by H or T (heads or tails) is ?, then we previously asserted that the parameters of the distribution for the number of H (or the number of T) is given byE(H) = n/2andσH = n/4 The formula for E(H) is rather obvious, but the formula for σH is not. This is a simple derivation of the σH formula for the case of n = 2.For the case of n = 2, there are 4 possible outcomes: (i) 2 successive H, (ii) 1 H followed by 1 T, (iii) 1 T followed by 1 H, or 2 successive Ts. The probability of each of these outcomes is ?. Note that there is 1 outcome with 2 heads, 2 outcomes with 1 head, and 1 outcome with no heads. The expected number of H is given by E(H) = n/2 = 2/2 = 1. The variance is given by σH2 = (2 – 1)2 (1/4) + 2(1 – 1)2 (1/4) + (0 – 1)2 (1/4) = ? + 0 + ? = ? σH = 1/2 = n/4 since n = 2.A general proof for any value of n, see Chapter 5 of your text: “binomial probability distribution.” The general proof requires the binomial expansion formula, but the proof above should give you an intuitive understanding of the process. ................
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