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2. [4 points) Let X1,..., Xn be a random sample from a distribution with mean li and variance o; and let Y1,...,Yn be a rando

HI I NEED JUSTIFICATIONS, PLEASE HELP ME UNDERSTAND THANKS!! :)

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Answer #1

Answer a.

This is based on Central Limit Theorem which states that "For large sample sizes, the distribution of sample mean is approximately normal irrespective of the distribution of the random variables".

Therorefore, here since the distribution is unknown and the sample sizes are large so the approximate confidence interval for \boldsymbol{\mu_x - \mu_y} will be based on quantiles of Normal Distribution.

Answer b.

This is based on the fact on which the t-distribution is based. The following assumptions must be satisfied in order to use t-distribution for Difference in Means:

1. The Population from which the samples are drawn must be Normal.

2. Population Variances must be equal.

3. Population Variances must be unknown.

Here, since population variances are equal but unknown and the population is Normal, so we can obtain the exact confidence interval of Difference in Means \boldsymbol{\mu_x - \mu_y} using t-distribution.

Answer c.

This is because there is no information on the sample sizes and also if sample sizes were to be assumed large, then too we would have obatined an approximate confidence interval for Difference in Means based on Central Limit Theorem. For large n, Binomial do not tend to t-distribution

Hence, Binomial Distribution cannot be approximated by t-distribution.

Answer d.

This depends on the application of Central Limit Theorem which says that for large n the Binomial Distribution tends to Normal DIstribution.

So the approximate Confidence Interval would be based on quantiles of Normal DIstribution.

Note: Wherever Central Limit Theorem is applied, we will always obtain Approximate Normal Distribution and hence approximate Confidence Intervals.

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