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3. (a) You wish to study the difference in life expectancy between chimpanzees in the wild and in captivity. Over a long period of time you observe 20 randomly chosen Chimps in the wild, and 25 in captivity. You note that for the observed Chimps in the wild, the lifetimes have a mean 20.3 years with a standard deviation of 1.3 years, and for those in captivity the mean is 22.3 years, with a standard deviation of 1.2 years. Assuming normality and equal variances, i. Construct a 95% confidence interval for the difference in average lifetimes 12 marks ii. Is their evidence to support the belief that the animals in captivity live 5 marks] of the two populations. longer on average?

math   Statistics-And-Probability   
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We have, ni 20, пг-25 X1-20.3, X2 22.3 S1 1.3, S2 1.2 (subscript 1 for wild chimps and subscript 2 for captivity) The (1-a) 100% confidence interval for difference in means is given by tox is two tailed critical t value for a 1-0.95 0.05 with 2025-2-43 degrees n1+n2 2 of freedom Using -T.INV.2T(0.05,43) we can calculate two tailed critical t value which is 2.016692199 Hence, 95% confidence interval for the difference in average lifetimes is given by 1.32 1.22 25 20.3- 22.3 + 2.0167 2 0.7602 (-2 - 0.7602,-2 0.7602) (-2.7602,-1.2398) ii. Both the confidence limits are negative ie. we are 95% sure that difference of population averages is negative μ1-u2 < 0 ie. μ1 < μ2 i.e. average lifetime of wild Chimps is less than average lifetime of chimps in captivity so we can support the belief that the animals in captivity live longer on average

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