Can someone order these from least to greatest please?
2/9,21%,0.21,11/50

98% confidence interval level estimate of the mean amount of mercury in the population is [0.28 , 1.18].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 98% confidence interval for the population mean is given by;

                        P.Q. =  ~

where, = sample mean amount =  = 0.73

            s = sample standard deviation = =  0.381

           n = sample size = 7

            =  population mean amount of mercury

Here for constructing 98% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 98% confidence interval for the population mean, is ;

P(-3.143 < < 3.143) = 0.98  {As the critical value of t at 6 degree

                                       of freedom are -3.143 & 3.143 with P = 1%}  

P(-3.143 < < 3.143) = 0.98

P( < < ) = 0.98

P( < < ) = 0.98

98% confidence interval for = [ , ]

                             = [ , ]

                             = [0.28 , 1.18]

Therefore, 98% confidence interval level estimate of the mean amount of mercury in the population is [0.28 , 1.18].

The interpretation of the above confidence interval is that we are 98% confident that the mean amount of mercury in the population will lie between 0.28 and 1.18.