# In a survey of 35 adult Americans, it was found that the mean age (in years) that people would like to live to is 87.9 with a standard deviation of 15.5. A) Explain how a large sample size is...

In a survey of 35 adult Americans, it was found that the mean age (in years) that people would like to live to is 87.9 with a standard deviation of 15.5.

A) Explain how a large sample size is useful in constructing a confidence interval for the mean age the people would like to live.

B. Construct and interpret a 95% confidence interval for the mean age that people would like to live.

C. How many American adults would need to be surveyed to estimate the mean age that people would like to live to within 2 years with 95% confidence?

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A) A large sample size is necessary in order to apply the Central Limit Theorem and assume Normality.

B) The mean of the ages the individuals would like to live was found to be 87.9 and the standard deviation about this mean was found to be 15.5. Therefore, assuming Normality, a 95% confidence interval for the mean desired age people would like to live is

87.9 +/- 1.96(15.5) = (57.52,118.28)

where 1.96 is the 95th percentile of a standard Normal distribution, ie`Phi^(-1)(0.95)`.

Upon repeated surveys of the same size, we expect that the 95% confidence interval so constructed (mean +/- 1.96sd) would contain the true unknown mean desired age for people to live 95% of the time.

C) The standard deviation about the mean in this sample was found to be 15.5. This is equal to an estimate of the sqrt(variance/n), ie sqrt(V[X])/n = sqrt(V[X])/35.

To be within 2 years with 95% confidence, we require the width of the 95% confidence interval to be <4, that is that 1.96sd < 2 or sd < 1.02.

Since the sd estimates sqrt(V[X])/n, based on the results of this current survey we would expect to need to survey at least sqrt(V[X])/1.02 = (15.5 x 35)/1.02 = 531.65 people, ie n = 532 people.

**A) A large sample size is necessary in order to be able to appeal to the Central Limit Theorem and assume that the sample is Normally distributed about the mean with standard deviation `sigma/n` where `sigma^2 = V[X]`**.

**B) 95% confidence interval is given by (57.52,118.28). Upon repeated surveys, we would expect a 95% confidence interval constructed in this way to contain the true unknown mean age 95% of the time.**

**C) Taking the 95th percentile of a standard Normal to be 1.96, we would require a survey size of at least 532 people. Taking it to be 2, we would require a survey size of at least 543 people.**