Statistics and Probability

Question

You intend to estimate a population mean μμ with the following sample.

53.5
44.2
38.2
45
48.1
42.9
42.2
49.6
41.5
28.3

You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

80% C.I. = 

Answer 1

Thank you for taking the time to post your question!

The general form for computing a confidence interval is

mean +/- t* (standard deviation / sqrt(sample size))

so it’s just a matter of locating what all of those values would be for the parameters of this question and then applying the formula!

using a calculator i find that mean = 43.35, standard deviation = 6.8803, sample size = 10, and then the critical t value for this 80% level is 1.383

putting it all together yields

43.35 +/- 1.383 x 6.8803 / sqrt(10)

=40.34 to 46.36 for the 80% confidence interval