Answer to Question #204787 in Statistics and Probability for KLAUSXMİKE

Question

If a 90% confidence interval for the mean is found to be (1.25,4.75) using a sample of n=100, find the 95% confidence interval for the mean. a. Is this interval narrower or wider? Why? b. In general, explain how the sample size and the confidence level affect the width of a confidence interval. 

Answer 1

Assume that the population standard deviation is known.

$CI=\bar X\pm E$

for the 90% CI, $E=\frac{4.75-1.25}{2}=1.75$

And $\bar X=1.25+1.75=3$

But $E=z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

For 90% confidence level, $z=1.645$

Thus,$1.75=1.645(\frac{\sigma}{\sqrt{100}})$ solving for sigma,

$\sigma=10.64$

a. 95% CI

For 95% CI

$Z_{\frac{\alpha}{2}}=1.96$

$E=1.96(\frac{10.64}{\sqrt{100}})=2.085$

$CI=3\pm2.085=(0.915,5.085)$

95% CI is wider than 90% CI.

b.

Increasing sample size decreases the width of the confidence interval since as sample size increases, sample mean approaches the population mean. however, increasing confidence level widens the width of the confidence interval since high confidence occurs when one is less precise.