Answer to Question #299108 in Statistics and Probability for mww

Question

A study was conducted to determine the average salary of a teacher. With this, 81 teachers were invited as samples. It was known that the mean salary of these sample teachers is Php 24,000.00 with a standard deviation of Php 1,000.00. Construct a 90% confidence interval estimate for the average salary of all teachers.

Answer 1

$\bar x=24000\\s=1000\\\alpha=0.1\\n=81$

A 90% confidence interval  for the average salary of all teachers is given as,

$CI=\bar x\pm t_{{\alpha\over2},n-1}{s\over\sqrt{n}}=\bar x\pm t_{0.05,80}{s\over\sqrt{n}}$

From the t distribution table, $t_{0.05,80}= 1.664125$

Therefore, the 90% confidence interval is,

$CI=24000\pm 1.664125\times{1000\over\sqrt{81}}=24000\pm 184.902778$

Which is given as,

$(23815.09, \space 24184.90)$