You intend to estimate a population mean mu with the following sample. 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.You believe the population is normally distributed. Find the 99.9% 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).

A 99.9% confidence interval for the population mean is [22.31, 59.91] .

Step-by-step explanation:

We are given the following sample values;

X = 45.3, 42.3, 53, 49, 15.2, 52.3, 45.6, 39.6, 39.4, 16.1, 54.4.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                       P.Q.  =    ~  

where, = sample mean =  = 41.11

             s = sample standard deviation =  =  13.59

             n = sample size = 11

Here for constructing a 99.9% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 99.9% confidence interval for the population mean, is;

P(-4.587 < < 4.587) = 0.999  {As the critical value of t at 10 degrees of

                                                freedom are -4.587 & 4.587 with P = 0.05%}    

P(-4.587 < < 4.587) = 0.999

P( < < ) = 0.999  

P( < < ) = 0.999

99.9% confidence interval for = [ , ]

                                       = [ , ]

                                       = [22.31, 59.91]

Therefore, a 99.9% confidence interval for the population mean is [22.31, 59.91] .

where is the figure