Question

1.     A random sample of 15 fence posts from a garden center was found to have a mean length of 1.84 m. The population standard deviation for these posts is taken to be 17cm.

(i) Use the t-distribution to find a symmetric 95% confidence interval for the mean length of such posts.

(ii) Interpret the 95% confidence interval found in part (i) above.

Answer 1

The formula for confidence interval is

$m{\scriptscriptstyle *}=m±Cr*\frac {\sigma} {\sqrt {n}}$ , where $m{\scriptscriptstyle *}$ - true mean value, m - mean value obtained from data, Cr - critical value depends on confidence level, $\sigma$ - population standard deviation(sample if population is unknown), n - sample size

In the given case we have to use two-sided t-value with n - 1 = 14 degrees of freedom as critical value(most likely it is said due to the small sample size), then

$P(T(14)>Cr)={\frac {1+0.95} 2}=0.975\implies Cr=2.145$ . So,

$m±Cr*\frac {\sigma} {\sqrt {n}}=1.84±2.145*{\frac {0.17} {\sqrt {15}}}=1.84±0.094$

The 95% confidence interval for population mean is (1.84 - 0.094, 1.84 + 0.094) = (1.746, 1.934)

(ii) The obtained result means that, based on the given data, we can estimate with 0.95 probability that the population mean lies between 1.746 and 1.934