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Stat. Question- Sample T test


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Stat. Question: Sample T test

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When do children in the UK develop preferences for brand-name products? In a recent study, marketers showed children identical pictures of athletic shoes. One picture was labeled FW, and one was labeled Run. The children were asked to evaluate the shoes based on their appearancen, quality, price, prestige,... 顯示更多 When do children in the UK develop preferences for brand-name products? In a recent study, marketers showed children identical pictures of athletic shoes. One picture was labeled FW, and one was labeled Run. The children were asked to evaluate the shoes based on their appearancen, quality, price, prestige, favorableness, and preference for owning. A score from 2 (highest product evaluation possible) to -2 (lowest product evaluation possible) was recorded for each child. The following table reports the results of the study: Age 8: a) FW: Sample Size 28 / Sample Mean: 0.87 / Sample Standard Deviation: 0.97 b) Run: Sample Size 21/ Sample Mean: 0.85 / Sample Standard Deviation: 1.04 Age12: a) FW: Sample size: 42/ SMean: 0.89/ SD: 1.04 b) Run: SSize: 44/ Smean: 0.09/ SD: 1.07 Age 16: a) FW: SSize: 38/ SMean: 0.48/ SD: 0.82 b) Run: SSize: 35/ SMean: -0.30 / SD:0.93 Q1: Conduct a pooled-variance t test for the difference between the two means for each of the three age groups. Use a level of sign. of 0.05 Q2: Wht assumptions are needed to conduct the tests in Q1? Q3: Write a brief summary of your findings

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Q1. Age 8: s = √{[(n1-1)*s1^2+(n2-1)*s2^2]/(n1+n2-2)} = √{[(28-1)*(0.97)^2+(21-1)*(1.04)^2]/(28+21-2)} = √(47.0363/47) = √(1.0008) = 1.0004 H0: μ1-μ2 = 0 H1: μ1-μ2 not = 0 t = [(X1_bar-X2_bar)-(μ1-μ2)]/[s*√(1/n1+1/n2)] = (0.87-0.85)/[1.0004*√(1/28+1/21)] = 0.02/0.289 = 0.07 Since the t-statistic is small, we do NOT reject H0. (p-value > 0.05) Age 12: s = √{[(42-1)*(1.04)^2+(44-1)*(1.07)^2]/(42+44-2)} = √(93.5763/84) = √(1.114) = 1.06 H0: μ1-μ2 = 0 H1: μ1-μ2 not = 0 t = (0.89-0.09)/[1.06*√(1/42+1/44)] = 0.8/0.228 = 3.51 Since the t-statistic is large, we reject H0. (p-value < 0.01) Age 16: s = √{[(38-1)*(0.82)^2+(35-1)*(0.93)^2]/(38+35-2)} = √(54.2854/71) = √(0.7646) = 0.8744 H0: μ1-μ2 = 0 H1: μ1-μ2 not = 0 t = (0.48+0.30)/[0.8744*√(1/38+1/35)] = 0.78/0.205 = 3.81 Since the t-statistic is large, we reject H0. (p-value < 0.01) Q2. Assumption: 2 populations originally are normally distributed. population variances are equal.(less important) Q3. We notice that the difference in sample mean for age 8 group is only 0.02 but at least 0.7 for the other 2 groups. Therefore, we can conclude that there is NO difference between the two means for age 8 children. However, the two means start to be different as we increase the age of a particular group.

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