dizxindlert7
2022-09-12
Answered

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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London Maldonado

Answered 2022-09-13
Author has **13** answers

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.

Answer:

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

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.

Answer:

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

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