1.Consider monthly household income in a large developing country. It is a continuous variable. Itsdistribution is skewed towards right. Suppose that we repeatedly sample from the households inthat country. Each time, we choose a sample of 1,500 households. We ask them how much theyearn, we record their responses, and we compute the mean household income for the entiresample. We, then, repeat this exercise 500 times.a)The obtained sample means are going to be skewed towards right; just like the originaldistribution. The mean of this sampling distribution is equal to the population mean forhousehold income, and the standard deviation of this sampling distribution is equal to thepopulation standard deviation.b)The obtained sample means are going to be distributed normally. The mean of thissampling distribution is equal to the population mean for household income, and thestandard deviation of this sampling distribution is equal to the population standarddeviation.c)The obtained sample means are going to be distributed normally. The mean of thissampling distribution is equal to the population mean for household income. However,the standard deviation of this sampling distribution is much smaller than the populationstandard deviation for household income.d)The obtained sample means are going to be distributed normally. The mean of thissampling distribution is equal to the population mean for household income. However,the standard deviation of this sampling distribution is much larger than the populationstandard deviation for household income.

2.It is impossible to repeat the above sampling for 500 times. It is too expensive. At best, we canrely on one sample. Here is the sample in hand: sample size (n)=900, sample mean (à µ?0)=$868,and sample standard deviation (S)=$450. We also know thatProb.(tâ‰¤-1.96)=2.5% when degreeof freedom (d.f.) is equal to 899. Given this information, we are 95% confident that thepopulation mean for monthly household income in this country:

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