naivlingr

2020-12-28

Is statistical inference intuitive to babies? In other words, are babies able to generalize from sample to population? In this study,1 8-month-old infants watched someone draw a sample of five balls from an opaque box. Each sample consisted of four balls of one color (red or white) and one ball of the other color. After observing the sample, the side of the box was lifted so the infants could see all of the balls inside (the population). Some boxes had an “expected” population, with balls in the same color proportions as the sample, while other boxes had an “unexpected” population, with balls in the opposite color proportion from the sample. Babies looked at the unexpected populations for an average of 9.9 seconds ($$sd = 4.5$$ seconds) and the expected populations for an average of 7.5 seconds ($$sd = 4.2$$ seconds). The sample size in each group was 20, and you may assume the data in each group are reasonably normally distributed. Is this convincing evidence that babies look longer at the unexpected population, suggesting that they make inferences about the population from the sample? Let group 1 and group 2 be the time spent looking at the unexpected and expected populations, respectively. A) Calculate the relevant sample statistic. Enter the exact answer. Sample statistic: _____

B) Calculate the t-statistic. Round your answer to two decimal places. t-statistic = ___________

C) Find the p-value. Round your answer to three decimal places. p-value =__________

Theodore Schwartz

From the given information , for group 1(unexpected): ${\stackrel{―}{x}}_{1}=9.9,{s}_{1}=4.5,n=20,$ for group 2(expected): ${\stackrel{―}{x}}_{2}=7.5,{s}_{2}=4.2,$ and ${n}_{2}=20.$ Null hypothesis ${H}_{0}:{\mu }_{1}={\mu }_{2}$ Alternative hypothesis ${H}_{\alpha }:{\mu }_{1}>{\mu }_{2}$

(A) Sample statistic ${\mu }_{\left({x}_{1}-{x}_{2}\right)}={\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}=9.9-7.5=2.4$

(B) The test statistic ${t}_{c\alpha 1}\frac{{\stackrel{―}{x}}_{1}-{\stackrel{―}{x}}_{2}}{\sqrt{\frac{{S}_{1}^{2}}{{n}_{1}}+\frac{{S}_{2}^{2}}{{n}_{2}}}}=\frac{9.9-7.5}{\frac{{\left(4.5\right)}^{2}}{20}+\frac{{\left(4.2\right)}^{2}}{20}}=1.7437$

Degrees of freedom $=\left({n}_{1}+{n}_{2}-2\right)=38$

(C) $P-value=P\left(t>1.7437\right)=0.045,$

$P-value=0.045$ less than the level of significance $\alpha =0.05,$ we reject the null hypothesis.

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