Answer to Question #101691 in Statistics and Probability for louis

Question

Thirty-five elderly people from Sarasota, Florida were randomly sampled to determine the difference between their reported height and actual height. The mean difference was 3 inches with a standard deviation of 2.4. Find a 90% confidence interval for the true mean difference between reported and actual heights for the elderly in Sarasota, Florida. Can we generalize this finding to all people in the United States? Why or why not?

Answer 1

The provided sample mean is $\bar X=3$ and the population standard deviation is $\sigma=2.4.$ The size of the sample is $n=35$ and the required confidence level is $90\ \%.$

Based on the provided information, the critical z-value for $\alpha=0.1$ is $z_c=1.645.$

The 90% confidence for the population mean $\mu$ is computed using the following expression

$$CI=(\bar X-z_c\times{\sigma \over \sqrt{n}},\ \bar X+z_c\times{\sigma \over \sqrt{n}})$$

Therefore, based on the information provided, the 90 % confidence for the population mean $\mu$ is

$$CI=(3-1.645\times{2.4 \over \sqrt{35}},\ 3+1.645\times{2.4 \over \sqrt{35}})$$

$$=(3-0.6673,\ 3+0.6673)$$

$$=(2.3327,\ 3.6673)$$

A 90% confidence interval is a range of values that you can be 90% certain contains the true mean of the population. This is not the same as a range that contains 90% of the values.

Hence we cannot generalize this finding to all people in the United States.