# Identify the distribution (normal, Student t, chi-square) that should be used in each of the fo llowing situations. If none of the three distributions can be used, what other method could be used? In constructing a confidence interval estimate of p, you have 1200 survey respondents and 5% of them answered "yes" to the first question.

Identify the distribution (normal, Student t, chi-square) that should be used in each of the fo llowing situations. If none of the three distributions can be used, what other method could be used? In constructing a confidence interval estimate of p, you have 1200 survey respondents and 5% of them answered "yes" to the first question.
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IN GENERAL
Normal distribution: confidence interval for the mean $\mu$ with $\sigma$ known (sampling distribution of the sample mean should be approximately normal)
Normal distribution: confidence interval the proportion p (number of successes and number of failures should both be at least 5).
Student t distribution: confidence interval for the mean $\mu$ with $\sigma$ unknown (sampling distribution of the sample mean should be approximately normal)
Chi-square distribution: confidence interval for the standard deviation $\sigma$ or the variance ${\sigma }^{2}$ (population distribution should be normal)
If none of the three distributions can be used, then we can use the bootstrap method.
n=1200
p=5%=0.05
We should use the normal distribution, because we are interested in a confidence interval for the population proportion p.
The requirement of at least 5 successes and at least 5 failures is also satisfied, because there are np=1200(0.05)=60 successes and n(1-p)=1200(1-0.05)=1140 failures.
Result:
Normal distribution