Question

Construct a 90% confidence interval for the population mean, . assume the population has a normal distribution. a sample of 15 randomly selected students has a grade point average of 2.86 with a standard deviation of 0.78

Answer 1

`CI=\{2.5053,3.2147\}`

Explanation:

Assuming the conditions for constructing a t-confidence interval are true, we use the formula
`\displaystyle CI=\bar{x}\pm t^*\biggr((s)/(√(n))\biggr)` where our sample mean is
`\bar{x}=2.86`, our sample standard deviation is
`s=0.78`, our sample size is
`n=15`, and a 90% confidence level is equivalent to a critical value of
`t^*=1.7613`,

`\displaystyle CI=\bar{x}\pm z^*\biggr((s)/(√(n))\biggr)\\\\CI=2.86\pm 1.761\biggr((0.78)/(√(15))\biggr)\\\\CI=2.86\pm0.3547\\\\CI=\{2.5053,3.2147\}`

Hence, we are 90% confident that the true population mean of a student's grade point average is between 2.5053 and 3.2147