Question

Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Find a 90% confidence interval for the true (population) mean of statistics exam scores.

Answer 1

Answer:

( 67.18, 68.82)

Step-by-step explanation:

Let $\mu$ be the true (population) mean of statistics exam scores. We have a large random sample of n = 36 scores with a sample mean (sample mean score) of $\bar{x} = 68$. We know that the population standard deviation is $\sigma = 3$. A pivotal quantity is $(\bar{x}-\mu)/(3/\sqrt{36}) = (\bar{x}-\mu)/(3/6) = (68-\mu)/(1/2)$ which is approximately normally distributed and $P(-z_{0.05}\leq(68-\mu)/(1/2)\leq z_{0.05}) = 0.90$ where $-z_{0.05} =-1.6449$, and so, $P(-1.6449\leq(68-\mu)/(1/2)\leq 1.6449) = 0.90$. Therefore the 90% confidence interval is (68-(1/2)(1.6449), 68+(1/2)(1.6449)), i.e., ( 67.18, 68.82)