Use the following values from a student’s t-distribution to establish a 95% confidence interval for the population mean given a sample size of 10, a sample mean of 6.25, and a sample standard deviation of 12. Assume that the population from which the sample is drawn is normally distributed and that the population variance is not known.
|Degrees of Freedom|p = 0.10|p = 0.05|p = 0.025|p = 0.01|
|9 |1.383|1.833|2.262|2.821|
|10 |1.372|1.812|2.228|2.764|
|11 |1.363|1.796|2.201|2.718|
The 95% confidence interval is closest to a:
- lower bound of −2.20 and an upper bound of 14.70.
- lower bound of −0.71 and an upper bound of 13.21.
- lower bound of −2.33 and an upper bound of 14.83.
In this problem they used the t-statistic of 2.262 which is the .025 level. I’m confused what made them divide by two since we have no idea whether or not it’s a one or two tailed test. Sorry if the formatting is off, I did my best. Thanks for the help.